(5x^2-7x+3)=(2x^2+8x+1)

Solution for (5x^2-7x+3)=(2x^2+8x+1) equation:

(5x^2-7x+3)=(2x^2+8x+1)

We move all terms to the left:

(5x^2-7x+3)-((2x^2+8x+1))=0

We get rid of parentheses

5x^2-7x-((2x^2+8x+1))+3=0


We calculate terms in parentheses: -((2x^2+8x+1)), so:

(2x^2+8x+1)

We get rid of parentheses

2x^2+8x+1

Back to the equation:

-(2x^2+8x+1)


We get rid of parentheses

5x^2-2x^2-7x-8x-1+3=0

We add all the numbers together, and all the variables

3x^2-15x+2=0

a = 3; b = -15; c = +2;
Δ = b2-4ac
Δ = -152-4·3·2
Δ = 201
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-15)-\sqrt{201}}{2*3}=\frac{15-\sqrt{201}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-15)+\sqrt{201}}{2*3}=\frac{15+\sqrt{201}}{6}
$