What are the solutions to the equation 4d^2+15d-4=0 ?

The solutions to the equation 4d^2+15d-4=0 are d= 1/4 ,d=-4

Find the zeros of 4d^2+15d-4=0

The zeros of 4d^2+15d-4=0 are d= 1/4 ,d=-4

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4 d^{2} + 15 d - 4 = 0

d = \frac{1}{4}, d = - 4

Decimal Notation

d = 0.25, d = - 4

Solution steps

4 d^{2} + 15 d - 4 = 0

Solve with the quadratic formula

d_{1, 2} = \frac{- 15 \pm 1 5 ^{2} - 4 \cdot 4 ( - 4 )}{2 \cdot 4}

1 5^{2} - 4 \cdot 4 (- 4) = 17

d_{1, 2} = \frac{- 15 \pm 17}{2 \cdot 4}

Separate the solutions

d_{1} = \frac{- 15 + 17}{2 \cdot 4}, d_{2} = \frac{- 15 - 17}{2 \cdot 4}

d = \frac{- 15 + 17}{2 \cdot 4} : \frac{1}{4}

d = \frac{- 15 - 17}{2 \cdot 4} : - 4

The solutions to the quadratic equation are:

d = \frac{1}{4}, d = - 4

4 r^{2} - 7 = 29

\frac{5 7}{35}

2 x + 5 \geq - 7

2 a^{- \frac{1}{2}}

C = \frac{25 ( 0.6 )}{100 - 0.6}

What are the solutions to the equation 4d^2+15d-4=0 ?
The solutions to the equation 4d^2+15d-4=0 are d= 1/4 ,d=-4
Find the zeros of 4d^2+15d-4=0
The zeros of 4d^2+15d-4=0 are d= 1/4 ,d=-4