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25+40x-5x^2=0
a = -5; b = 40; c = +25;
Δ = b2-4ac
Δ = 402-4·(-5)·25
Δ = 2100
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{2100}=\sqrt{100*21}=\sqrt{100}*\sqrt{21}=10\sqrt{21}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-10\sqrt{21}}{2*-5}=\frac{-40-10\sqrt{21}}{-10} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+10\sqrt{21}}{2*-5}=\frac{-40+10\sqrt{21}}{-10} $
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