3x^2+10=-17x

Solution for 3x^2+10=-17x equation:

3x^2+10=-17x

We move all terms to the left:

3x^2+10-(-17x)=0

We get rid of parentheses

3x^2+17x+10=0

a = 3; b = 17; c = +10;
Δ = b2-4ac
Δ = 172-4·3·10
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{169}=13$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(17)-13}{2*3}=\frac{-30}{6}
=-5
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(17)+13}{2*3}=\frac{-4}{6}
=-2/3
$