3e2+25=133

Solution for 3e2+25=133 equation:

3e^2+25=133

We move all terms to the left:

3e^2+25-(133)=0

We add all the numbers together, and all the variables

3e^2-108=0

a = 3; b = 0; c = -108;
Δ = b2-4ac
Δ = 02-4·3·(-108)
Δ = 1296
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$e_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$e_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1296}=36$
$e_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-36}{2*3}=\frac{-36}{6}
=-6
$
$e_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+36}{2*3}=\frac{36}{6}
=6
$