5w^2+22w-21=0

Solution for 5w^2+22w-21=0 equation:

5w^2+22w-21=0

a = 5; b = 22; c = -21;
Δ = b2-4ac
Δ = 222-4·5·(-21)
Δ = 904
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{904}=\sqrt{4*226}=\sqrt{4}*\sqrt{226}=2\sqrt{226}$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(22)-2\sqrt{226}}{2*5}=\frac{-22-2\sqrt{226}}{10}
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(22)+2\sqrt{226}}{2*5}=\frac{-22+2\sqrt{226}}{10}
$