Solve \[{{x}^{2}}-10x=-24?\]

Hint: In order to solve the given equation \[{{x}^{2}}-10x=-24\], first of all we will write all the terms of the given equation on left hand side and then equate it with zero \[\left( 0 \right)\]. After converting the given equation into standard form we will find the value of \[x\] using factorization method it will factorize the equation into linear equation and simplifying these linear factors we will get our required answer.

Complete step by step solution:
Given equation can be written as \[{{x}^{2}}-10x+24=0\] ..............(i)
Which is a standard form of quadratic equation.
Now, we will find two numbers that multiply to give \[24\] and add to give \[-10\].
So, we can write \[24\] as \[-6\times -4\] and also by adding \[6\] and \[4\] we will get it as \[10\].
So, equation \[1\] can be written as \[{{x}^{2}}-6x-4x+24=0\]
On comparing equation (i) with standard form of quadratic equation \[a{{x}^{2}}+bx+c=0\] we get
\[ac\] as \[24\times 1=24\] and \[b\] is \[-10\]
So, we want two numbers that multiply together to make \[24\] and add up to \[-10\].
In fact \[-6\] and \[-4\] do that (as \[-6x-4=24\] and \[-6+\left( -4 \right)=-10\])
Thus, we will rewrite the middle term with those numbers. That is we will rewrite \[-10x\] with \[-6x\] and \[-4x\].
\[\Rightarrow {{x}^{2}}-6x-4x+24=0\]
Now, factor the first two and last two terms respectively. The first two terms that is \[{{x}^{2}}-6x\] factors into \[x\left( x-6 \right)\] and the last two terms that is \[-4x+24\] factors into \[-4\left( x-6 \right)\].
So we get,
\[x\left( x-6 \right)-4\left( x-6 \right)=0\] .............(ii)
It is clearly visible that \[\left( x-6 \right)\] is common to both the terms
\[\therefore \]equation (ii) can be written as
\[\left( x-6 \right)\left( x-4 \right)=0\] (taking \[x-6\] as common)
\[\Rightarrow x-6=0\] or \[x-4=0\]
 \[\Rightarrow x=6\] or \[x=4\]
\[\Rightarrow x=6,4\] is the required answer.

Note: If we have give \[A\] quadratic equation in standard form that is \[a{{x}^{2}}+bx+c=0\]; \[a,b\] and \[c\] can have any value, except that a can not be equal to zero.
For finding roots of the quadratic equation (or to solve) we will follow the steps mentioned below.
• Find two numbers that multiply to give ac (in other words a times \[c\]) and add to give \[b.\]
• Rewrite the middle terms with those numbers
• Factor the first two and last two terms separately.
• If we have done the above steps correctly then two new terms should have a clearly visible common factor.
• Now equate it with zero and we get our required roots.