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a^2+4a-38=0
a = 1; b = 4; c = -38;
Δ = b2-4ac
Δ = 42-4·1·(-38)
Δ = 168
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{168}=\sqrt{4*42}=\sqrt{4}*\sqrt{42}=2\sqrt{42}$$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-2\sqrt{42}}{2*1}=\frac{-4-2\sqrt{42}}{2} $$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+2\sqrt{42}}{2*1}=\frac{-4+2\sqrt{42}}{2} $
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