(8x+9)=(3x^2+x-1)

Solution for (8x+9)=(3x^2+x-1) equation:

(8x+9)=(3x^2+x-1)

We move all terms to the left:

(8x+9)-((3x^2+x-1))=0

We get rid of parentheses

8x-((3x^2+x-1))+9=0


We calculate terms in parentheses: -((3x^2+x-1)), so:

(3x^2+x-1)

We get rid of parentheses

3x^2+x-1

Back to the equation:

-(3x^2+x-1)


We get rid of parentheses

-3x^2+8x-x+1+9=0

We add all the numbers together, and all the variables

-3x^2+7x+10=0

a = -3; b = 7; c = +10;
Δ = b2-4ac
Δ = 72-4·(-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{-(7)-13}{2*-3}=\frac{-20}{-6}
=3+1/3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+13}{2*-3}=\frac{6}{-6}
=-1
$