(2x+3)(4x-5)=-15

Solution for (2x+3)(4x-5)=-15 equation:

(2x+3)(4x-5)=-15

We move all terms to the left:

(2x+3)(4x-5)-(-15)=0

We add all the numbers together, and all the variables

(2x+3)(4x-5)+15=0

We multiply parentheses ..

(+8x^2-10x+12x-15)+15=0

We get rid of parentheses

8x^2-10x+12x-15+15=0

We add all the numbers together, and all the variables

8x^2+2x=0

a = 8; b = 2; c = 0;
Δ = b2-4ac
Δ = 22-4·8·0
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2}{2*8}=\frac{-4}{16}
=-1/4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2}{2*8}=\frac{0}{16}
=0
$