0=10x^2+38x+13

Solution for 0=10x^2+38x+13 equation:

0=10x^2+38x+13

We move all terms to the left:

0-(10x^2+38x+13)=0

We add all the numbers together, and all the variables

-(10x^2+38x+13)=0

We get rid of parentheses

-10x^2-38x-13=0

a = -10; b = -38; c = -13;
Δ = b2-4ac
Δ = -382-4·(-10)·(-13)
Δ = 924
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{924}=\sqrt{4*231}=\sqrt{4}*\sqrt{231}=2\sqrt{231}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-38)-2\sqrt{231}}{2*-10}=\frac{38-2\sqrt{231}}{-20}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-38)+2\sqrt{231}}{2*-10}=\frac{38+2\sqrt{231}}{-20}
$