4x(x-7)+5x(2x+3)=13(x^2-x+2)-10

Solution for 4x(x-7)+5x(2x+3)=13(x^2-x+2)-10 equation:

4x(x-7)+5x(2x+3)=13(x^2-x+2)-10

We move all terms to the left:

4x(x-7)+5x(2x+3)-(13(x^2-x+2)-10)=0

We multiply parentheses

4x^2+10x^2-28x+15x-(13(x^2-x+2)-10)=0


We calculate terms in parentheses: -(13(x^2-x+2)-10), so:

13(x^2-x+2)-10

We multiply parentheses

13x^2-13x+26-10

We add all the numbers together, and all the variables

13x^2-13x+16

Back to the equation:

-(13x^2-13x+16)


We add all the numbers together, and all the variables

14x^2-13x-(13x^2-13x+16)=0

We get rid of parentheses

14x^2-13x^2-13x+13x-16=0

We add all the numbers together, and all the variables

x^2-16=0

a = 1; b = 0; c = -16;
Δ = b2-4ac
Δ = 02-4·1·(-16)
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8}{2*1}=\frac{-8}{2}
=-4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8}{2*1}=\frac{8}{2}
=4
$