3h2+17h=-10

Solution for 3h2+17h=-10 equation:

3h^2+17h=-10

We move all terms to the left:

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

We add all the numbers together, and all the variables

3h^2+17h+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:
$h_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$h_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

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