3y^2-y^2+(-10)=0

Solution for 3y^2-y^2+(-10)=0 equation:

3y^2-y^2+(-10)=0

We add all the numbers together, and all the variables

2y^2-10=0

a = 2; b = 0; c = -10;
Δ = b2-4ac
Δ = 02-4·2·(-10)
Δ = 80
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{80}=\sqrt{16*5}=\sqrt{16}*\sqrt{5}=4\sqrt{5}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{5}}{2*2}=\frac{0-4\sqrt{5}}{4}
=-\frac{4\sqrt{5}}{4}
=-\sqrt{5}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{5}}{2*2}=\frac{0+4\sqrt{5}}{4}
=\frac{4\sqrt{5}}{4}
=\sqrt{5}
$