E=(x-5)(x+1)+(x-5)(2x-3)

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

=(E-5)(E+1)+(E-5)(2E-3)

We move all terms to the left:

-((E-5)(E+1)+(E-5)(2E-3))=0

We multiply parentheses ..

-((+E^2+E-5E-5)+(E-5)(2E-3))=0


We calculate terms in parentheses: -((+E^2+E-5E-5)+(E-5)(2E-3)), so:

(+E^2+E-5E-5)+(E-5)(2E-3)

We get rid of parentheses

E^2+E-5E+(E-5)(2E-3)-5

We multiply parentheses ..

E^2+(+2E^2-3E-10E+15)+E-5E-5

We add all the numbers together, and all the variables

E^2+(+2E^2-3E-10E+15)-4E-5

We get rid of parentheses

E^2+2E^2-3E-10E-4E+15-5

We add all the numbers together, and all the variables

3E^2-17E+10

Back to the equation:

-(3E^2-17E+10)


We get rid of parentheses

-3E^2+17E-10=0

a = -3; b = 17; c = -10;
Δ = b2-4ac
Δ = 172-4·(-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:
$E_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$E_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{169}=13$
$E_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(17)-13}{2*-3}=\frac{-30}{-6}
=+5
$
$E_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(17)+13}{2*-3}=\frac{-4}{-6}
=2/3
$