23x2+19x-12=5688

Solution for 23x2+19x-12=5688 equation:

23x^2+19x-12=5688

We move all terms to the left:

23x^2+19x-12-(5688)=0

We add all the numbers together, and all the variables

23x^2+19x-5700=0

a = 23; b = 19; c = -5700;
Δ = b2-4ac
Δ = 192-4·23·(-5700)
Δ = 524761
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{524761}}{2*23}=\frac{-19-\sqrt{524761}}{46}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+\sqrt{524761}}{2*23}=\frac{-19+\sqrt{524761}}{46}
$