(x+4)(4x-5)=70

Solution for (x+4)(4x-5)=70 equation:

(x+4)(4x-5)=70

We move all terms to the left:

(x+4)(4x-5)-(70)=0

We multiply parentheses ..

(+4x^2-5x+16x-20)-70=0

We get rid of parentheses

4x^2-5x+16x-20-70=0

We add all the numbers together, and all the variables

4x^2+11x-90=0

a = 4; b = 11; c = -90;
Δ = b2-4ac
Δ = 112-4·4·(-90)
Δ = 1561
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(11)-\sqrt{1561}}{2*4}=\frac{-11-\sqrt{1561}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+\sqrt{1561}}{2*4}=\frac{-11+\sqrt{1561}}{8}
$