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4k^2+10k-36=0
a = 4; b = 10; c = -36;
Δ = b2-4ac
Δ = 102-4·4·(-36)
Δ = 676
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{676}=26$$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-26}{2*4}=\frac{-36}{8} =-4+1/2 $$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+26}{2*4}=\frac{16}{8} =2 $
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