3r^2+6r-7=0

Solution for 3r^2+6r-7=0 equation:

3r^2+6r-7=0

a = 3; b = 6; c = -7;
Δ = b2-4ac
Δ = 62-4·3·(-7)
Δ = 120
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:

$\sqrt{\Delta}=\sqrt{120}=\sqrt{4*30}=\sqrt{4}*\sqrt{30}=2\sqrt{30}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-2\sqrt{30}}{2*3}=\frac{-6-2\sqrt{30}}{6}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+2\sqrt{30}}{2*3}=\frac{-6+2\sqrt{30}}{6}
$