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23-3x^2=5+3x
We move all terms to the left:
23-3x^2-(5+3x)=0
We add all the numbers together, and all the variables
-3x^2-(3x+5)+23=0
We get rid of parentheses
-3x^2-3x-5+23=0
We add all the numbers together, and all the variables
-3x^2-3x+18=0
a = -3; b = -3; c = +18;
Δ = b2-4ac
Δ = -32-4·(-3)·18
Δ = 225
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{225}=15$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-15}{2*-3}=\frac{-12}{-6} =+2 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+15}{2*-3}=\frac{18}{-6} =-3 $
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