(x-3)(4x-5)=15

Solution for (x-3)(4x-5)=15 equation:

(x-3)(4x-5)=15

We move all terms to the left:

(x-3)(4x-5)-(15)=0

We multiply parentheses ..

(+4x^2-5x-12x+15)-15=0

We get rid of parentheses

4x^2-5x-12x+15-15=0

We add all the numbers together, and all the variables

4x^2-17x=0

a = 4; b = -17; c = 0;
Δ = b2-4ac
Δ = -172-4·4·0
Δ = 289
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{289}=17$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-17)-17}{2*4}=\frac{0}{8}
=0
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-17)+17}{2*4}=\frac{34}{8}
=4+1/4
$