23x+2x^2=16

Solution for 23x+2x^2=16 equation:

23x+2x^2=16

We move all terms to the left:

23x+2x^2-(16)=0

a = 2; b = 23; c = -16;
Δ = b2-4ac
Δ = 232-4·2·(-16)
Δ = 657
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:

$\sqrt{\Delta}=\sqrt{657}=\sqrt{9*73}=\sqrt{9}*\sqrt{73}=3\sqrt{73}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(23)-3\sqrt{73}}{2*2}=\frac{-23-3\sqrt{73}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(23)+3\sqrt{73}}{2*2}=\frac{-23+3\sqrt{73}}{4}
$