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-5t^2+18=-3+18
We move all terms to the left:
-5t^2+18-(-3+18)=0
We add all the numbers together, and all the variables
-5t^2+18-15=0
We add all the numbers together, and all the variables
-5t^2+3=0
a = -5; b = 0; c = +3;
Δ = b2-4ac
Δ = 02-4·(-5)·3
Δ = 60
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{60}=\sqrt{4*15}=\sqrt{4}*\sqrt{15}=2\sqrt{15}$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{15}}{2*-5}=\frac{0-2\sqrt{15}}{-10} =-\frac{2\sqrt{15}}{-10} =-\frac{\sqrt{15}}{-5} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{15}}{2*-5}=\frac{0+2\sqrt{15}}{-10} =\frac{2\sqrt{15}}{-10} =\frac{\sqrt{15}}{-5} $
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