3x^2+(4x^2+20)=90

Solution for 3x^2+(4x^2+20)=90 equation:

3x^2+(4x^2+20)=90

We move all terms to the left:

3x^2+(4x^2+20)-(90)=0

We get rid of parentheses

3x^2+4x^2+20-90=0

We add all the numbers together, and all the variables

7x^2-70=0

a = 7; b = 0; c = -70;
Δ = b2-4ac
Δ = 02-4·7·(-70)
Δ = 1960
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:

$\sqrt{\Delta}=\sqrt{1960}=\sqrt{196*10}=\sqrt{196}*\sqrt{10}=14\sqrt{10}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-14\sqrt{10}}{2*7}=\frac{0-14\sqrt{10}}{14}
=-\frac{14\sqrt{10}}{14}
=-\sqrt{10}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+14\sqrt{10}}{2*7}=\frac{0+14\sqrt{10}}{14}
=\frac{14\sqrt{10}}{14}
=\sqrt{10}
$