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-x^2+3x+5=1
We move all terms to the left:
-x^2+3x+5-(1)=0
We add all the numbers together, and all the variables
-1x^2+3x+4=0
a = -1; b = 3; c = +4;
Δ = b2-4ac
Δ = 32-4·(-1)·4
Δ = 25
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{25}=5$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-5}{2*-1}=\frac{-8}{-2} =+4 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+5}{2*-1}=\frac{2}{-2} =-1 $
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