5(x^2+2)=2x^2+41

Solution for 5(x^2+2)=2x^2+41 equation:

5(x^2+2)=2x^2+41

We move all terms to the left:

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

We multiply parentheses

5x^2-(2x^2+41)+10=0

We get rid of parentheses

5x^2-2x^2-41+10=0

We add all the numbers together, and all the variables

3x^2-31=0

a = 3; b = 0; c = -31;
Δ = b2-4ac
Δ = 02-4·3·(-31)
Δ = 372
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:

$\sqrt{\Delta}=\sqrt{372}=\sqrt{4*93}=\sqrt{4}*\sqrt{93}=2\sqrt{93}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{93}}{2*3}=\frac{0-2\sqrt{93}}{6}
=-\frac{2\sqrt{93}}{6}
=-\frac{\sqrt{93}}{3}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{93}}{2*3}=\frac{0+2\sqrt{93}}{6}
=\frac{2\sqrt{93}}{6}
=\frac{\sqrt{93}}{3}
$