3x^2+14x=25

Solution for 3x^2+14x=25 equation:

3x^2+14x=25

We move all terms to the left:

3x^2+14x-(25)=0

a = 3; b = 14; c = -25;
Δ = b2-4ac
Δ = 142-4·3·(-25)
Δ = 496
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:

$\sqrt{\Delta}=\sqrt{496}=\sqrt{16*31}=\sqrt{16}*\sqrt{31}=4\sqrt{31}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(14)-4\sqrt{31}}{2*3}=\frac{-14-4\sqrt{31}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+4\sqrt{31}}{2*3}=\frac{-14+4\sqrt{31}}{6}
$