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5x^2=28x+49
We move all terms to the left:
5x^2-(28x+49)=0
We get rid of parentheses
5x^2-28x-49=0
a = 5; b = -28; c = -49;
Δ = b2-4ac
Δ = -282-4·5·(-49)
Δ = 1764
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{1764}=42$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-28)-42}{2*5}=\frac{-14}{10} =-1+2/5 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-28)+42}{2*5}=\frac{70}{10} =7 $
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