2x+3+79x^2=25

Solution for 2x+3+79x^2=25 equation:

2x+3+79x^2=25

We move all terms to the left:

2x+3+79x^2-(25)=0

We add all the numbers together, and all the variables

79x^2+2x-22=0

a = 79; b = 2; c = -22;
Δ = b2-4ac
Δ = 22-4·79·(-22)
Δ = 6956
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{6956}=\sqrt{4*1739}=\sqrt{4}*\sqrt{1739}=2\sqrt{1739}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{1739}}{2*79}=\frac{-2-2\sqrt{1739}}{158}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{1739}}{2*79}=\frac{-2+2\sqrt{1739}}{158}
$