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b(2b-6)=10
We move all terms to the left:
b(2b-6)-(10)=0
We multiply parentheses
2b^2-6b-10=0
a = 2; b = -6; c = -10;
Δ = b2-4ac
Δ = -62-4·2·(-10)
Δ = 116
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{116}=\sqrt{4*29}=\sqrt{4}*\sqrt{29}=2\sqrt{29}$$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-2\sqrt{29}}{2*2}=\frac{6-2\sqrt{29}}{4} $$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+2\sqrt{29}}{2*2}=\frac{6+2\sqrt{29}}{4} $
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