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2x^2+3.5x-5=0
a = 2; b = 3.5; c = -5;
Δ = b2-4ac
Δ = 3.52-4·2·(-5)
Δ = 52.25
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3.5)-\sqrt{52.25}}{2*2}=\frac{-3.5-\sqrt{52.25}}{4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3.5)+\sqrt{52.25}}{2*2}=\frac{-3.5+\sqrt{52.25}}{4} $
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