x2+20x-10=13x-5

Solution for x2+20x-10=13x-5 equation:

x2+20x-10=13x-5

We move all terms to the left:

x2+20x-10-(13x-5)=0

We add all the numbers together, and all the variables

x^2+20x-(13x-5)-10=0

We get rid of parentheses

x^2+20x-13x+5-10=0

We add all the numbers together, and all the variables

x^2+7x-5=0

a = 1; b = 7; c = -5;
Δ = b2-4ac
Δ = 72-4·1·(-5)
Δ = 69
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-\sqrt{69}}{2*1}=\frac{-7-\sqrt{69}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+\sqrt{69}}{2*1}=\frac{-7+\sqrt{69}}{2}
$