23x+42=5x^2

Solution for 23x+42=5x^2 equation:

23x+42=5x^2

We move all terms to the left:

23x+42-(5x^2)=0

determiningTheFunctionDomain
-5x^2+23x+42=0

a = -5; b = 23; c = +42;
Δ = b2-4ac
Δ = 232-4·(-5)·42
Δ = 1369
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1369}=37$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(23)-37}{2*-5}=\frac{-60}{-10}
=+6
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(23)+37}{2*-5}=\frac{14}{-10}
=-1+2/5
$