21x^2+21x=96x-36

Solution for 21x^2+21x=96x-36 equation:

21x^2+21x=96x-36

We move all terms to the left:

21x^2+21x-(96x-36)=0

We get rid of parentheses

21x^2+21x-96x+36=0

We add all the numbers together, and all the variables

21x^2-75x+36=0

a = 21; b = -75; c = +36;
Δ = b2-4ac
Δ = -752-4·21·36
Δ = 2601
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2601}=51$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-75)-51}{2*21}=\frac{24}{42}
=4/7
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-75)+51}{2*21}=\frac{126}{42}
=3
$