What are the solutions to the equation 4d^2+9d-7=22 ?

The solutions to the equation 4d^2+9d-7=22 are d=(-9+sqrt(545))/8 ,d=(-9-sqrt(545))/8

Find the zeros of 4d^2+9d-7=22

The zeros of 4d^2+9d-7=22 are d=(-9+sqrt(545))/8 ,d=(-9-sqrt(545))/8

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4 d^{2} + 9 d - 7 = 22

d = \frac{- 9 + 545}{8}, d = \frac{- 9 - 545}{8}

Decimal Notation

d = 1.79315 \dots, d = - 4.04315 \dots

Solution steps

4 d^{2} + 9 d - 7 = 22

Move

22

to the left side

4 d^{2} + 9 d - 29 = 0

Solve with the quadratic formula

d_{1, 2} = \frac{- 9 \pm 9 ^{2} - 4 \cdot 4 ( - 29 )}{2 \cdot 4}

9^{2} - 4 \cdot 4 (- 29) = 545

d_{1, 2} = \frac{- 9 \pm 545}{2 \cdot 4}

Separate the solutions

d_{1} = \frac{- 9 + 545}{2 \cdot 4}, d_{2} = \frac{- 9 - 545}{2 \cdot 4}

d = \frac{- 9 + 545}{2 \cdot 4} : \frac{- 9 + 545}{8}

d = \frac{- 9 - 545}{2 \cdot 4} : \frac{- 9 - 545}{8}

The solutions to the quadratic equation are:

d = \frac{- 9 + 545}{8}, d = \frac{- 9 - 545}{8}

- x^{2} + 10 x

x, y^{2} - x y + x^{2} = 8

(x - 3)^{2} = 9 x

- 6 x^{2} + 4 x + 5 = 0

\frac{7}{5} x + \frac{14}{5} = x^{2} + \frac{2}{5} x - \frac{16}{5}

What are the solutions to the equation 4d^2+9d-7=22 ?
The solutions to the equation 4d^2+9d-7=22 are d=(-9+sqrt(545))/8 ,d=(-9-sqrt(545))/8
Find the zeros of 4d^2+9d-7=22
The zeros of 4d^2+9d-7=22 are d=(-9+sqrt(545))/8 ,d=(-9-sqrt(545))/8