2x^2+23x=25

Solution for 2x^2+23x=25 equation:

2x^2+23x=25

We move all terms to the left:

2x^2+23x-(25)=0

a = 2; b = 23; c = -25;
Δ = b2-4ac
Δ = 232-4·2·(-25)
Δ = 729
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{729}=27$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(23)-27}{2*2}=\frac{-50}{4}
=-12+1/2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(23)+27}{2*2}=\frac{4}{4}
=1
$