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4x(3x+40)=30
We move all terms to the left:
4x(3x+40)-(30)=0
We multiply parentheses
12x^2+160x-30=0
a = 12; b = 160; c = -30;
Δ = b2-4ac
Δ = 1602-4·12·(-30)
Δ = 27040
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{27040}=\sqrt{2704*10}=\sqrt{2704}*\sqrt{10}=52\sqrt{10}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(160)-52\sqrt{10}}{2*12}=\frac{-160-52\sqrt{10}}{24} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(160)+52\sqrt{10}}{2*12}=\frac{-160+52\sqrt{10}}{24} $
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