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-2.5x^2+20x+5=0
a = -2.5; b = 20; c = +5;
Δ = b2-4ac
Δ = 202-4·(-2.5)·5
Δ = 450
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{450}=\sqrt{225*2}=\sqrt{225}*\sqrt{2}=15\sqrt{2}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-15\sqrt{2}}{2*-2.5}=\frac{-20-15\sqrt{2}}{-5} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+15\sqrt{2}}{2*-2.5}=\frac{-20+15\sqrt{2}}{-5} $
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