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5x^2-38x+48=0
a = 5; b = -38; c = +48;
Δ = b2-4ac
Δ = -382-4·5·48
Δ = 484
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{484}=22$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-38)-22}{2*5}=\frac{16}{10} =1+3/5 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-38)+22}{2*5}=\frac{60}{10} =6 $
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