26x^2+16x-64=0

Solution for 26x^2+16x-64=0 equation:

26x^2+16x-64=0

a = 26; b = 16; c = -64;
Δ = b2-4ac
Δ = 162-4·26·(-64)
Δ = 6912
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:

$\sqrt{\Delta}=\sqrt{6912}=\sqrt{2304*3}=\sqrt{2304}*\sqrt{3}=48\sqrt{3}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-48\sqrt{3}}{2*26}=\frac{-16-48\sqrt{3}}{52}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+48\sqrt{3}}{2*26}=\frac{-16+48\sqrt{3}}{52}
$