2r^2+14r-3=0

Solution for 2r^2+14r-3=0 equation:

2r^2+14r-3=0

a = 2; b = 14; c = -3;
Δ = b2-4ac
Δ = 142-4·2·(-3)
Δ = 220
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:

$\sqrt{\Delta}=\sqrt{220}=\sqrt{4*55}=\sqrt{4}*\sqrt{55}=2\sqrt{55}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(14)-2\sqrt{55}}{2*2}=\frac{-14-2\sqrt{55}}{4}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+2\sqrt{55}}{2*2}=\frac{-14+2\sqrt{55}}{4}
$