90=(7x-3)(4x+5)

Solution for 90=(7x-3)(4x+5) equation:

90=(7x-3)(4x+5)

We move all terms to the left:

90-((7x-3)(4x+5))=0

We multiply parentheses ..

-((+28x^2+35x-12x-15))+90=0


We calculate terms in parentheses: -((+28x^2+35x-12x-15)), so:

(+28x^2+35x-12x-15)

We get rid of parentheses

28x^2+35x-12x-15

We add all the numbers together, and all the variables

28x^2+23x-15

Back to the equation:

-(28x^2+23x-15)


We get rid of parentheses

-28x^2-23x+15+90=0

We add all the numbers together, and all the variables

-28x^2-23x+105=0

a = -28; b = -23; c = +105;
Δ = b2-4ac
Δ = -232-4·(-28)·105
Δ = 12289
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{-(-23)-\sqrt{12289}}{2*-28}=\frac{23-\sqrt{12289}}{-56}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-23)+\sqrt{12289}}{2*-28}=\frac{23+\sqrt{12289}}{-56}
$