2x^2=19x+10

Solution for 2x^2=19x+10 equation:

2x^2=19x+10

We move all terms to the left:

2x^2-(19x+10)=0

We get rid of parentheses

2x^2-19x-10=0

a = 2; b = -19; c = -10;
Δ = b2-4ac
Δ = -192-4·2·(-10)
Δ = 441
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{441}=21$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-19)-21}{2*2}=\frac{-2}{4}
=-1/2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-19)+21}{2*2}=\frac{40}{4}
=10
$