5x(2x+1)+6x-13=3(3x-2)

Solution for 5x(2x+1)+6x-13=3(3x-2) equation:

5x(2x+1)+6x-13=3(3x-2)

We move all terms to the left:

5x(2x+1)+6x-13-(3(3x-2))=0

We add all the numbers together, and all the variables

6x+5x(2x+1)-(3(3x-2))-13=0

We multiply parentheses

10x^2+6x+5x-(3(3x-2))-13=0


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

3(3x-2)

We multiply parentheses

9x-6

Back to the equation:

-(9x-6)


We add all the numbers together, and all the variables

10x^2+11x-(9x-6)-13=0

We get rid of parentheses

10x^2+11x-9x+6-13=0

We add all the numbers together, and all the variables

10x^2+2x-7=0

a = 10; b = 2; c = -7;
Δ = b2-4ac
Δ = 22-4·10·(-7)
Δ = 284
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{284}=\sqrt{4*71}=\sqrt{4}*\sqrt{71}=2\sqrt{71}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{71}}{2*10}=\frac{-2-2\sqrt{71}}{20}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{71}}{2*10}=\frac{-2+2\sqrt{71}}{20}
$