5u^2+41u=42=0

Solution for 5u^2+41u=42=0 equation:

5u^2+41u=42=0

We move all terms to the left:

5u^2+41u-(42)=0

a = 5; b = 41; c = -42;
Δ = b2-4ac
Δ = 412-4·5·(-42)
Δ = 2521
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(41)-\sqrt{2521}}{2*5}=\frac{-41-\sqrt{2521}}{10}
$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(41)+\sqrt{2521}}{2*5}=\frac{-41+\sqrt{2521}}{10}
$