8x^2+16x-41=0

Solution for 8x^2+16x-41=0 equation:

8x^2+16x-41=0

a = 8; b = 16; c = -41;
Δ = b2-4ac
Δ = 162-4·8·(-41)
Δ = 1568
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:

$\sqrt{\Delta}=\sqrt{1568}=\sqrt{784*2}=\sqrt{784}*\sqrt{2}=28\sqrt{2}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-28\sqrt{2}}{2*8}=\frac{-16-28\sqrt{2}}{16}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+28\sqrt{2}}{2*8}=\frac{-16+28\sqrt{2}}{16}
$