(3x-15)=(x^2-5x)

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

(3x-15)=(x^2-5x)

We move all terms to the left:

(3x-15)-((x^2-5x))=0

We get rid of parentheses

3x-((x^2-5x))-15=0


We calculate terms in parentheses: -((x^2-5x)), so:

(x^2-5x)

We get rid of parentheses

x^2-5x

Back to the equation:

-(x^2-5x)


We get rid of parentheses

-x^2+3x+5x-15=0

We add all the numbers together, and all the variables

-1x^2+8x-15=0

a = -1; b = 8; c = -15;
Δ = b2-4ac
Δ = 82-4·(-1)·(-15)
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{-(8)-2}{2*-1}=\frac{-10}{-2}
=+5
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+2}{2*-1}=\frac{-6}{-2}
=+3
$