3x^2+9x-10=18-8x

Solution for 3x^2+9x-10=18-8x equation:

3x^2+9x-10=18-8x

We move all terms to the left:

3x^2+9x-10-(18-8x)=0

We add all the numbers together, and all the variables

3x^2+9x-(-8x+18)-10=0

We get rid of parentheses

3x^2+9x+8x-18-10=0

We add all the numbers together, and all the variables

3x^2+17x-28=0

a = 3; b = 17; c = -28;
Δ = b2-4ac
Δ = 172-4·3·(-28)
Δ = 625
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{625}=25$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(17)-25}{2*3}=\frac{-42}{6}
=-7
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(17)+25}{2*3}=\frac{8}{6}
=1+1/3
$