5u^2+27u+10=0

Solution for 5u^2+27u+10=0 equation:

5u^2+27u+10=0

a = 5; b = 27; c = +10;
Δ = b2-4ac
Δ = 272-4·5·10
Δ = 529
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{529}=23$
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(27)-23}{2*5}=\frac{-50}{10}
=-5
$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(27)+23}{2*5}=\frac{-4}{10}
=-2/5
$