(X^2-5x)+(7x+20)=10x+5

Solution for (X^2-5x)+(7x+20)=10x+5 equation:

(X^2-5X)+(7X+20)=10X+5

We move all terms to the left:

(X^2-5X)+(7X+20)-(10X+5)=0

We get rid of parentheses

X^2-5X+7X-10X+20-5=0

We add all the numbers together, and all the variables

X^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{6}{2}
=3
$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+2}{2*1}=\frac{10}{2}
=5
$