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-3x^2=-45
We move all terms to the left:
-3x^2-(-45)=0
We add all the numbers together, and all the variables
-3x^2+45=0
a = -3; b = 0; c = +45;
Δ = b2-4ac
Δ = 02-4·(-3)·45
Δ = 540
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{540}=\sqrt{36*15}=\sqrt{36}*\sqrt{15}=6\sqrt{15}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-6\sqrt{15}}{2*-3}=\frac{0-6\sqrt{15}}{-6} =-\frac{6\sqrt{15}}{-6} =-\frac{\sqrt{15}}{-1} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+6\sqrt{15}}{2*-3}=\frac{0+6\sqrt{15}}{-6} =\frac{6\sqrt{15}}{-6} =\frac{\sqrt{15}}{-1} $
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