-4a^2+12a+17=0

Solution for -4a^2+12a+17=0 equation:

-4a^2+12a+17=0

a = -4; b = 12; c = +17;
Δ = b2-4ac
Δ = 122-4·(-4)·17
Δ = 416
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{416}=\sqrt{16*26}=\sqrt{16}*\sqrt{26}=4\sqrt{26}$
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-4\sqrt{26}}{2*-4}=\frac{-12-4\sqrt{26}}{-8}
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+4\sqrt{26}}{2*-4}=\frac{-12+4\sqrt{26}}{-8}
$