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0.5x^2-3x-27=0
a = 0.5; b = -3; c = -27;
Δ = b2-4ac
Δ = -32-4·0.5·(-27)
Δ = 63
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{63}=\sqrt{9*7}=\sqrt{9}*\sqrt{7}=3\sqrt{7}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-3\sqrt{7}}{2*0.5}=\frac{3-3\sqrt{7}}{1} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+3\sqrt{7}}{2*0.5}=\frac{3+3\sqrt{7}}{1} $
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