3x^2+10x-8-25=0

Solution for 3x^2+10x-8-25=0 equation:

3x^2+10x-8-25=0

We add all the numbers together, and all the variables

3x^2+10x-33=0

a = 3; b = 10; c = -33;
Δ = b2-4ac
Δ = 102-4·3·(-33)
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{-(10)-4\sqrt{31}}{2*3}=\frac{-10-4\sqrt{31}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+4\sqrt{31}}{2*3}=\frac{-10+4\sqrt{31}}{6}
$