(3u+1)(7u-5)=0

Solution for (3u+1)(7u-5)=0 equation:

(3u+1)(7u-5)=0

We multiply parentheses ..

(+21u^2-15u+7u-5)=0

We get rid of parentheses

21u^2-15u+7u-5=0

We add all the numbers together, and all the variables

21u^2-8u-5=0

a = 21; b = -8; c = -5;
Δ = b2-4ac
Δ = -82-4·21·(-5)
Δ = 484
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{484}=22$
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-22}{2*21}=\frac{-14}{42}
=-1/3
$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+22}{2*21}=\frac{30}{42}
=5/7
$