x^2-4x+61=10x+13

Solution for x^2-4x+61=10x+13 equation:

x^2-4x+61=10x+13

We move all terms to the left:

x^2-4x+61-(10x+13)=0

We get rid of parentheses

x^2-4x-10x-13+61=0

We add all the numbers together, and all the variables

x^2-14x+48=0

a = 1; b = -14; c = +48;
Δ = b2-4ac
Δ = -142-4·1·48
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{-(-14)-2}{2*1}=\frac{12}{2}
=6
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-14)+2}{2*1}=\frac{16}{2}
=8
$