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300+5x-x^2=0
We add all the numbers together, and all the variables
-1x^2+5x+300=0
a = -1; b = 5; c = +300;
Δ = b2-4ac
Δ = 52-4·(-1)·300
Δ = 1225
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{1225}=35$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-35}{2*-1}=\frac{-40}{-2} =+20 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+35}{2*-1}=\frac{30}{-2} =-15 $
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