3x^2+10x=400

Solution for 3x^2+10x=400 equation:

3x^2+10x=400

We move all terms to the left:

3x^2+10x-(400)=0

a = 3; b = 10; c = -400;
Δ = b2-4ac
Δ = 102-4·3·(-400)
Δ = 4900
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4900}=70$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-70}{2*3}=\frac{-80}{6}
=-13+1/3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+70}{2*3}=\frac{60}{6}
=10
$