(4x+7)(x-4)=20x+35

Solution for (4x+7)(x-4)=20x+35 equation:

(4x+7)(x-4)=20x+35

We move all terms to the left:

(4x+7)(x-4)-(20x+35)=0

We get rid of parentheses

(4x+7)(x-4)-20x-35=0

We multiply parentheses ..

(+4x^2-16x+7x-28)-20x-35=0

We get rid of parentheses

4x^2-16x+7x-20x-28-35=0

We add all the numbers together, and all the variables

4x^2-29x-63=0

a = 4; b = -29; c = -63;
Δ = b2-4ac
Δ = -292-4·4·(-63)
Δ = 1849
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1849}=43$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-29)-43}{2*4}=\frac{-14}{8}
=-1+3/4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-29)+43}{2*4}=\frac{72}{8}
=9
$