6(1-u^2)+7u-7=0

Solution for 6(1-u^2)+7u-7=0 equation:

6(1-u^2)+7u-7=0

We multiply parentheses

-6u^2+7u+6-7=0

We add all the numbers together, and all the variables

-6u^2+7u-1=0

a = -6; b = 7; c = -1;
Δ = b2-4ac
Δ = 72-4·(-6)·(-1)
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-5}{2*-6}=\frac{-12}{-12}
=1
$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+5}{2*-6}=\frac{-2}{-12}
=1/6
$