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w2+10w+3=0
We add all the numbers together, and all the variables
w^2+10w+3=0
a = 1; b = 10; c = +3;
Δ = b2-4ac
Δ = 102-4·1·3
Δ = 88
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}$$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{88}=\sqrt{4*22}=\sqrt{4}*\sqrt{22}=2\sqrt{22}$$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2\sqrt{22}}{2*1}=\frac{-10-2\sqrt{22}}{2} $$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2\sqrt{22}}{2*1}=\frac{-10+2\sqrt{22}}{2} $
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