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0.25x^2+4x+3=0
a = 0.25; b = 4; c = +3;
Δ = b2-4ac
Δ = 42-4·0.25·3
Δ = 13
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{-(4)-\sqrt{13}}{2*0.25}=\frac{-4-\sqrt{13}}{0.5} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+\sqrt{13}}{2*0.25}=\frac{-4+\sqrt{13}}{0.5} $
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