3x^2+4x^2+10=50

Solution for 3x^2+4x^2+10=50 equation:

3x^2+4x^2+10=50

We move all terms to the left:

3x^2+4x^2+10-(50)=0

We add all the numbers together, and all the variables

7x^2-40=0

a = 7; b = 0; c = -40;
Δ = b2-4ac
Δ = 02-4·7·(-40)
Δ = 1120
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:

$\sqrt{\Delta}=\sqrt{1120}=\sqrt{16*70}=\sqrt{16}*\sqrt{70}=4\sqrt{70}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{70}}{2*7}=\frac{0-4\sqrt{70}}{14}
=-\frac{4\sqrt{70}}{14}
=-\frac{2\sqrt{70}}{7}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{70}}{2*7}=\frac{0+4\sqrt{70}}{14}
=\frac{4\sqrt{70}}{14}
=\frac{2\sqrt{70}}{7}
$