16x^2+7x-23=0

Solution for 16x^2+7x-23=0 equation:

16x^2+7x-23=0

a = 16; b = 7; c = -23;
Δ = b2-4ac
Δ = 72-4·16·(-23)
Δ = 1521
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1521}=39$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-39}{2*16}=\frac{-46}{32}
=-1+7/16
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+39}{2*16}=\frac{32}{32}
=1
$