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