17x=3x^2+10

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

17x=3x^2+10

We move all terms to the left:

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