2c^2+6c^2-80=0

Solution for 2c^2+6c^2-80=0 equation:

2c^2+6c^2-80=0

We add all the numbers together, and all the variables

8c^2-80=0

a = 8; b = 0; c = -80;
Δ = b2-4ac
Δ = 02-4·8·(-80)
Δ = 2560
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{2560}=\sqrt{256*10}=\sqrt{256}*\sqrt{10}=16\sqrt{10}$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-16\sqrt{10}}{2*8}=\frac{0-16\sqrt{10}}{16}
=-\frac{16\sqrt{10}}{16}
=-\sqrt{10}
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+16\sqrt{10}}{2*8}=\frac{0+16\sqrt{10}}{16}
=\frac{16\sqrt{10}}{16}
=\sqrt{10}
$