(x^2+7x)+(-6x^2-3x=)

Solution for (x^2+7x)+(-6x^2-3x=) equation:

(x^2+7x)+(-6x^2-3x=)

We move all terms to the left:

(x^2+7x)+(-6x^2-3x-())=0

We get rid of parentheses

(-6x^2-3x-())+x^2+7x=0


We calculate terms in parentheses: +(-6x^2-3x-()), so:

-6x^2-3x-()

We add all the numbers together, and all the variables

-6x^2-3x

Back to the equation:

+(-6x^2-3x)


We add all the numbers together, and all the variables

x^2+(-6x^2-3x)+7x=0

We get rid of parentheses

x^2-6x^2-3x+7x=0

We add all the numbers together, and all the variables

-5x^2+4x=0

a = -5; b = 4; c = 0;
Δ = b2-4ac
Δ = 42-4·(-5)·0
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4}{2*-5}=\frac{-8}{-10}
=4/5
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4}{2*-5}=\frac{0}{-10}
=0
$