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x^2-4x-5=3x+3
We move all terms to the left:
x^2-4x-5-(3x+3)=0
We get rid of parentheses
x^2-4x-3x-3-5=0
We add all the numbers together, and all the variables
x^2-7x-8=0
a = 1; b = -7; c = -8;
Δ = b2-4ac
Δ = -72-4·1·(-8)
Δ = 81
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{81}=9$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-7)-9}{2*1}=\frac{-2}{2} =-1 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-7)+9}{2*1}=\frac{16}{2} =8 $
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