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5x^2-47x+48=0
a = 5; b = -47; c = +48;
Δ = b2-4ac
Δ = -472-4·5·48
Δ = 1249
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-47)-\sqrt{1249}}{2*5}=\frac{47-\sqrt{1249}}{10} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-47)+\sqrt{1249}}{2*5}=\frac{47+\sqrt{1249}}{10} $
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