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5x^2-40x+48=0
a = 5; b = -40; c = +48;
Δ = b2-4ac
Δ = -402-4·5·48
Δ = 640
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{640}=\sqrt{64*10}=\sqrt{64}*\sqrt{10}=8\sqrt{10}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-40)-8\sqrt{10}}{2*5}=\frac{40-8\sqrt{10}}{10} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-40)+8\sqrt{10}}{2*5}=\frac{40+8\sqrt{10}}{10} $
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