5x^2-7=3x^2+41

Solution for 5x^2-7=3x^2+41 equation:

5x^2-7=3x^2+41

We move all terms to the left:

5x^2-7-(3x^2+41)=0

We get rid of parentheses

5x^2-3x^2-41-7=0

We add all the numbers together, and all the variables

2x^2-48=0

a = 2; b = 0; c = -48;
Δ = b2-4ac
Δ = 02-4·2·(-48)
Δ = 384
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:

$\sqrt{\Delta}=\sqrt{384}=\sqrt{64*6}=\sqrt{64}*\sqrt{6}=8\sqrt{6}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8\sqrt{6}}{2*2}=\frac{0-8\sqrt{6}}{4}
=-\frac{8\sqrt{6}}{4}
=-2\sqrt{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{6}}{2*2}=\frac{0+8\sqrt{6}}{4}
=\frac{8\sqrt{6}}{4}
=2\sqrt{6}
$