9x^2+16x-3=0

Solution for 9x^2+16x-3=0 equation:

9x^2+16x-3=0

a = 9; b = 16; c = -3;
Δ = b2-4ac
Δ = 162-4·9·(-3)
Δ = 364
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:

$\sqrt{\Delta}=\sqrt{364}=\sqrt{4*91}=\sqrt{4}*\sqrt{91}=2\sqrt{91}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-2\sqrt{91}}{2*9}=\frac{-16-2\sqrt{91}}{18}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+2\sqrt{91}}{2*9}=\frac{-16+2\sqrt{91}}{18}
$