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10m^2-20m-48=0
a = 10; b = -20; c = -48;
Δ = b2-4ac
Δ = -202-4·10·(-48)
Δ = 2320
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{2320}=\sqrt{16*145}=\sqrt{16}*\sqrt{145}=4\sqrt{145}$$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-4\sqrt{145}}{2*10}=\frac{20-4\sqrt{145}}{20} $$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+4\sqrt{145}}{2*10}=\frac{20+4\sqrt{145}}{20} $
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