10x^2-23=4x^2+31

Solution for 10x^2-23=4x^2+31 equation:

10x^2-23=4x^2+31

We move all terms to the left:

10x^2-23-(4x^2+31)=0

We get rid of parentheses

10x^2-4x^2-31-23=0

We add all the numbers together, and all the variables

6x^2-54=0

a = 6; b = 0; c = -54;
Δ = b2-4ac
Δ = 02-4·6·(-54)
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{-(0)-36}{2*6}=\frac{-36}{12}
=-3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+36}{2*6}=\frac{36}{12}
=3
$