17x^2+50x+3=0

Solution for 17x^2+50x+3=0 equation:

17x^2+50x+3=0

a = 17; b = 50; c = +3;
Δ = b2-4ac
Δ = 502-4·17·3
Δ = 2296
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{2296}=\sqrt{4*574}=\sqrt{4}*\sqrt{574}=2\sqrt{574}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(50)-2\sqrt{574}}{2*17}=\frac{-50-2\sqrt{574}}{34}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(50)+2\sqrt{574}}{2*17}=\frac{-50+2\sqrt{574}}{34}
$