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0.25x^2+20x-195=0
a = 0.25; b = 20; c = -195;
Δ = b2-4ac
Δ = 202-4·0.25·(-195)
Δ = 595
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{-(20)-\sqrt{595}}{2*0.25}=\frac{-20-\sqrt{595}}{0.5} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+\sqrt{595}}{2*0.25}=\frac{-20+\sqrt{595}}{0.5} $
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