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(2x+5)x=300
We move all terms to the left:
(2x+5)x-(300)=0
We multiply parentheses
2x^2+5x-300=0
a = 2; b = 5; c = -300;
Δ = b2-4ac
Δ = 52-4·2·(-300)
Δ = 2425
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{2425}=\sqrt{25*97}=\sqrt{25}*\sqrt{97}=5\sqrt{97}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-5\sqrt{97}}{2*2}=\frac{-5-5\sqrt{97}}{4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+5\sqrt{97}}{2*2}=\frac{-5+5\sqrt{97}}{4} $
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