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2x(3x+20)=45
We move all terms to the left:
2x(3x+20)-(45)=0
We multiply parentheses
6x^2+40x-45=0
a = 6; b = 40; c = -45;
Δ = b2-4ac
Δ = 402-4·6·(-45)
Δ = 2680
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{2680}=\sqrt{4*670}=\sqrt{4}*\sqrt{670}=2\sqrt{670}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-2\sqrt{670}}{2*6}=\frac{-40-2\sqrt{670}}{12} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+2\sqrt{670}}{2*6}=\frac{-40+2\sqrt{670}}{12} $
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