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5=3.15r^2
We move all terms to the left:
5-(3.15r^2)=0
We get rid of parentheses
-3.15r^2+5=0
a = -3.15; b = 0; c = +5;
Δ = b2-4ac
Δ = 02-4·(-3.15)·5
Δ = 63
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{63}=\sqrt{9*7}=\sqrt{9}*\sqrt{7}=3\sqrt{7}$$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-3\sqrt{7}}{2*-3.15}=\frac{0-3\sqrt{7}}{-6.3} =-\frac{3\sqrt{7}}{-6.3} =-\frac{\sqrt{7}}{-2.1} $$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+3\sqrt{7}}{2*-3.15}=\frac{0+3\sqrt{7}}{-6.3} =\frac{3\sqrt{7}}{-6.3} =\frac{\sqrt{7}}{-2.1} $
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