What are the solutions to the equation 4k^2-22k-14=0 ?

The solutions to the equation 4k^2-22k-14=0 are k=(11+sqrt(177))/4 ,k=(11-sqrt(177))/4

Find the zeros of 4k^2-22k-14=0

The zeros of 4k^2-22k-14=0 are k=(11+sqrt(177))/4 ,k=(11-sqrt(177))/4

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4 k^{2} - 22 k - 14 = 0

k = \frac{11 + 177}{4}, k = \frac{11 - 177}{4}

Decimal Notation

k = 6.07603 \dots, k = - 0.57603 \dots

Solution steps

4 k^{2} - 22 k - 14 = 0

Solve with the quadratic formula

k_{1, 2} = \frac{- ( - 22 ) \pm ( - 22 ) ^{2} - 4 \cdot 4 ( - 14 )}{2 \cdot 4}

(- 22)^{2} - 4 \cdot 4 (- 14) = 2177

k_{1, 2} = \frac{- ( - 22 ) \pm 2 177}{2 \cdot 4}

Separate the solutions

k_{1} = \frac{- ( - 22 ) + 2 177}{2 \cdot 4}, k_{2} = \frac{- ( - 22 ) - 2 177}{2 \cdot 4}

k = \frac{- ( - 22 ) + 2 177}{2 \cdot 4} : \frac{11 + 177}{4}

k = \frac{- ( - 22 ) - 2 177}{2 \cdot 4} : \frac{11 - 177}{4}

The solutions to the quadratic equation are:

k = \frac{11 + 177}{4}, k = \frac{11 - 177}{4}

5^{2} + (117)^{2} = x^{2}

x^{2} - \frac{3}{2} x - \frac{16}{7} = 0

- a^{2} - 2 = 0

x (x - 4) - 2 x = - x (x + 1) + 21

121 x^{2} - 9 = 0

What are the solutions to the equation 4k^2-22k-14=0 ?
The solutions to the equation 4k^2-22k-14=0 are k=(11+sqrt(177))/4 ,k=(11-sqrt(177))/4
Find the zeros of 4k^2-22k-14=0
The zeros of 4k^2-22k-14=0 are k=(11+sqrt(177))/4 ,k=(11-sqrt(177))/4