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8r^2+14r-7=10r
We move all terms to the left:
8r^2+14r-7-(10r)=0
We add all the numbers together, and all the variables
8r^2+4r-7=0
a = 8; b = 4; c = -7;
Δ = b2-4ac
Δ = 42-4·8·(-7)
Δ = 240
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}$$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{240}=\sqrt{16*15}=\sqrt{16}*\sqrt{15}=4\sqrt{15}$$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4\sqrt{15}}{2*8}=\frac{-4-4\sqrt{15}}{16} $$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4\sqrt{15}}{2*8}=\frac{-4+4\sqrt{15}}{16} $
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