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-4+7-3+5=10R^2-7
We move all terms to the left:
-4+7-3+5-(10R^2-7)=0
We add all the numbers together, and all the variables
-(10R^2-7)+5=0
We get rid of parentheses
-10R^2+7+5=0
We add all the numbers together, and all the variables
-10R^2+12=0
a = -10; b = 0; c = +12;
Δ = b2-4ac
Δ = 02-4·(-10)·12
Δ = 480
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}$$R_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{480}=\sqrt{16*30}=\sqrt{16}*\sqrt{30}=4\sqrt{30}$$R_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{30}}{2*-10}=\frac{0-4\sqrt{30}}{-20} =-\frac{4\sqrt{30}}{-20} =-\frac{\sqrt{30}}{-5} $$R_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{30}}{2*-10}=\frac{0+4\sqrt{30}}{-20} =\frac{4\sqrt{30}}{-20} =\frac{\sqrt{30}}{-5} $
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