5w^2+51w+10=0

Solution for 5w^2+51w+10=0 equation:

5w^2+51w+10=0

a = 5; b = 51; c = +10;
Δ = b2-4ac
Δ = 512-4·5·10
Δ = 2401
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2401}=49$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(51)-49}{2*5}=\frac{-100}{10}
=-10
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(51)+49}{2*5}=\frac{-2}{10}
=-1/5
$