7x^2-20x=32+8x+4x^

Solution for 7x^2-20x=32+8x+4x^ equation:

7x^2-20x=32+8x+4x^

We move all terms to the left:

7x^2-20x-(32+8x+4x^)=0

We get rid of parentheses

7x^2-20x-8x-4x^-32=0

We add all the numbers together, and all the variables

7x^2-32x-32=0

a = 7; b = -32; c = -32;
Δ = b2-4ac
Δ = -322-4·7·(-32)
Δ = 1920
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{1920}=\sqrt{64*30}=\sqrt{64}*\sqrt{30}=8\sqrt{30}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-32)-8\sqrt{30}}{2*7}=\frac{32-8\sqrt{30}}{14}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-32)+8\sqrt{30}}{2*7}=\frac{32+8\sqrt{30}}{14}
$