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-1+6k-((k-3)(5k+3))=0
We multiply parentheses ..
-((+5k^2+3k-15k-9))+6k-1=0
We calculate terms in parentheses: -((+5k^2+3k-15k-9)), so:We add all the numbers together, and all the variables
(+5k^2+3k-15k-9)
We get rid of parentheses
5k^2+3k-15k-9
We add all the numbers together, and all the variables
5k^2-12k-9
Back to the equation:
-(5k^2-12k-9)
6k-(5k^2-12k-9)-1=0
We get rid of parentheses
-5k^2+6k+12k+9-1=0
We add all the numbers together, and all the variables
-5k^2+18k+8=0
a = -5; b = 18; c = +8;
Δ = b2-4ac
Δ = 182-4·(-5)·8
Δ = 484
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{484}=22$$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-22}{2*-5}=\frac{-40}{-10} =+4 $$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+22}{2*-5}=\frac{4}{-10} =-2/5 $
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