47x^2+38x-17=0

Solution for 47x^2+38x-17=0 equation:

47x^2+38x-17=0

a = 47; b = 38; c = -17;
Δ = b2-4ac
Δ = 382-4·47·(-17)
Δ = 4640
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:

$\sqrt{\Delta}=\sqrt{4640}=\sqrt{16*290}=\sqrt{16}*\sqrt{290}=4\sqrt{290}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(38)-4\sqrt{290}}{2*47}=\frac{-38-4\sqrt{290}}{94}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(38)+4\sqrt{290}}{2*47}=\frac{-38+4\sqrt{290}}{94}
$