6u(5u+2)=-3u(3u-3)

Solution for 6u(5u+2)=-3u(3u-3) equation:

6u(5u+2)=-3u(3u-3)

We move all terms to the left:

6u(5u+2)-(-3u(3u-3))=0

We multiply parentheses

30u^2+12u-(-3u(3u-3))=0


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

-3u(3u-3)

We multiply parentheses

-9u^2+9u

Back to the equation:

-(-9u^2+9u)


We get rid of parentheses

30u^2+9u^2-9u+12u=0

We add all the numbers together, and all the variables

39u^2+3u=0

a = 39; b = 3; c = 0;
Δ = b2-4ac
Δ = 32-4·39·0
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9}=3$
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-3}{2*39}=\frac{-6}{78}
=-1/13
$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+3}{2*39}=\frac{0}{78}
=0
$