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2w^2+4w-420=0
a = 2; b = 4; c = -420;
Δ = b2-4ac
Δ = 42-4·2·(-420)
Δ = 3376
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}$$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{3376}=\sqrt{16*211}=\sqrt{16}*\sqrt{211}=4\sqrt{211}$$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4\sqrt{211}}{2*2}=\frac{-4-4\sqrt{211}}{4} $$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4\sqrt{211}}{2*2}=\frac{-4+4\sqrt{211}}{4} $
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