2x^2-22x+38=-10

Solution for 2x^2-22x+38=-10 equation:

2x^2-22x+38=-10

We move all terms to the left:

2x^2-22x+38-(-10)=0

We add all the numbers together, and all the variables

2x^2-22x+48=0

a = 2; b = -22; c = +48;
Δ = b2-4ac
Δ = -222-4·2·48
Δ = 100
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{100}=10$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-22)-10}{2*2}=\frac{12}{4}
=3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-22)+10}{2*2}=\frac{32}{4}
=8
$