-20x-40=-5x(3x+13)

Solution for -20x-40=-5x(3x+13) equation:

-20x-40=-5x(3x+13)

We move all terms to the left:

-20x-40-(-5x(3x+13))=0


We calculate terms in parentheses: -(-5x(3x+13)), so:

-5x(3x+13)

We multiply parentheses

-15x^2-65x

Back to the equation:

-(-15x^2-65x)


We get rid of parentheses

15x^2+65x-20x-40=0

We add all the numbers together, and all the variables

15x^2+45x-40=0

a = 15; b = 45; c = -40;
Δ = b2-4ac
Δ = 452-4·15·(-40)
Δ = 4425
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{4425}=\sqrt{25*177}=\sqrt{25}*\sqrt{177}=5\sqrt{177}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(45)-5\sqrt{177}}{2*15}=\frac{-45-5\sqrt{177}}{30}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(45)+5\sqrt{177}}{2*15}=\frac{-45+5\sqrt{177}}{30}
$