5x^2-3x=(7x+5x)+(2x^2+16)

Solution for 5x^2-3x=(7x+5x)+(2x^2+16) equation:

5x^2-3x=(7x+5x)+(2x^2+16)

We move all terms to the left:

5x^2-3x-((7x+5x)+(2x^2+16))=0

We add all the numbers together, and all the variables

5x^2-3x-((+12x)+(2x^2+16))=0


We calculate terms in parentheses: -((+12x)+(2x^2+16)), so:

(+12x)+(2x^2+16)

We get rid of parentheses

2x^2+12x+16

Back to the equation:

-(2x^2+12x+16)


We get rid of parentheses

5x^2-2x^2-3x-12x-16=0

We add all the numbers together, and all the variables

3x^2-15x-16=0

a = 3; b = -15; c = -16;
Δ = b2-4ac
Δ = -152-4·3·(-16)
Δ = 417
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{417}}{2*3}=\frac{15-\sqrt{417}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-15)+\sqrt{417}}{2*3}=\frac{15+\sqrt{417}}{6}
$