6r^2+53-70=0

Solution for 6r^2+53-70=0 equation:

6r^2+53-70=0

We add all the numbers together, and all the variables

6r^2-17=0

a = 6; b = 0; c = -17;
Δ = b2-4ac
Δ = 02-4·6·(-17)
Δ = 408
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{408}=\sqrt{4*102}=\sqrt{4}*\sqrt{102}=2\sqrt{102}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{102}}{2*6}=\frac{0-2\sqrt{102}}{12}
=-\frac{2\sqrt{102}}{12}
=-\frac{\sqrt{102}}{6}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{102}}{2*6}=\frac{0+2\sqrt{102}}{12}
=\frac{2\sqrt{102}}{12}
=\frac{\sqrt{102}}{6}
$