23x^2+19x-12=0

Solution for 23x^2+19x-12=0 equation:

23x^2+19x-12=0

a = 23; b = 19; c = -12;
Δ = b2-4ac
Δ = 192-4·23·(-12)
Δ = 1465
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(19)-\sqrt{1465}}{2*23}=\frac{-19-\sqrt{1465}}{46}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+\sqrt{1465}}{2*23}=\frac{-19+\sqrt{1465}}{46}
$