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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 = 19
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{19}}{2*-0.25}=\frac{4-\sqrt{19}}{-0.5} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+\sqrt{19}}{2*-0.25}=\frac{4+\sqrt{19}}{-0.5} $
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