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x(2x+3)=152
We move all terms to the left:
x(2x+3)-(152)=0
We multiply parentheses
2x^2+3x-152=0
a = 2; b = 3; c = -152;
Δ = b2-4ac
Δ = 32-4·2·(-152)
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$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-35}{2*2}=\frac{-38}{4} =-9+1/2 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+35}{2*2}=\frac{32}{4} =8 $
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