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0.25x^2+15x-195=0
a = 0.25; b = 15; c = -195;
Δ = b2-4ac
Δ = 152-4·0.25·(-195)
Δ = 420
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{420}=\sqrt{4*105}=\sqrt{4}*\sqrt{105}=2\sqrt{105}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(15)-2\sqrt{105}}{2*0.25}=\frac{-15-2\sqrt{105}}{0.5} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(15)+2\sqrt{105}}{2*0.25}=\frac{-15+2\sqrt{105}}{0.5} $
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