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x(2x+3)-127=0
We multiply parentheses
2x^2+3x-127=0
a = 2; b = 3; c = -127;
Δ = b2-4ac
Δ = 32-4·2·(-127)
Δ = 1025
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1025}=\sqrt{25*41}=\sqrt{25}*\sqrt{41}=5\sqrt{41}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-5\sqrt{41}}{2*2}=\frac{-3-5\sqrt{41}}{4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+5\sqrt{41}}{2*2}=\frac{-3+5\sqrt{41}}{4} $
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