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.5(3x^2-40)=166
We move all terms to the left:
.5(3x^2-40)-(166)=0
We multiply parentheses
1.5x^2-20-166=0
We add all the numbers together, and all the variables
1.5x^2-186=0
a = 1.5; b = 0; c = -186;
Δ = b2-4ac
Δ = 02-4·1.5·(-186)
Δ = 1116
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1116}=\sqrt{36*31}=\sqrt{36}*\sqrt{31}=6\sqrt{31}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-6\sqrt{31}}{2*1.5}=\frac{0-6\sqrt{31}}{3} =-\frac{6\sqrt{31}}{3} =-2\sqrt{31} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+6\sqrt{31}}{2*1.5}=\frac{0+6\sqrt{31}}{3} =\frac{6\sqrt{31}}{3} =2\sqrt{31} $
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