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2x^2+35x-5.5=0
a = 2; b = 35; c = -5.5;
Δ = b2-4ac
Δ = 352-4·2·(-5.5)
Δ = 1269
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1269}=\sqrt{9*141}=\sqrt{9}*\sqrt{141}=3\sqrt{141}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(35)-3\sqrt{141}}{2*2}=\frac{-35-3\sqrt{141}}{4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(35)+3\sqrt{141}}{2*2}=\frac{-35+3\sqrt{141}}{4} $
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