3a^2+16a+17=0

Solution for 3a^2+16a+17=0 equation:

3a^2+16a+17=0

a = 3; b = 16; c = +17;
Δ = b2-4ac
Δ = 162-4·3·17
Δ = 52
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:

$\sqrt{\Delta}=\sqrt{52}=\sqrt{4*13}=\sqrt{4}*\sqrt{13}=2\sqrt{13}$
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-2\sqrt{13}}{2*3}=\frac{-16-2\sqrt{13}}{6}
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+2\sqrt{13}}{2*3}=\frac{-16+2\sqrt{13}}{6}
$