6x^2+12x-38=10

Solution for 6x^2+12x-38=10 equation:

6x^2+12x-38=10

We move all terms to the left:

6x^2+12x-38-(10)=0

We add all the numbers together, and all the variables

6x^2+12x-48=0

a = 6; b = 12; c = -48;
Δ = b2-4ac
Δ = 122-4·6·(-48)
Δ = 1296
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1296}=36$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-36}{2*6}=\frac{-48}{12}
=-4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+36}{2*6}=\frac{24}{12}
=2
$