What are the solutions to the equation d^2+3d-7=0 ?

The solutions to the equation d^2+3d-7=0 are d=(-3+sqrt(37))/2 ,d=(-3-sqrt(37))/2

Find the zeros of d^2+3d-7=0

The zeros of d^2+3d-7=0 are d=(-3+sqrt(37))/2 ,d=(-3-sqrt(37))/2

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d^{2} + 3 d - 7 = 0

d = \frac{- 3 + 37}{2}, d = \frac{- 3 - 37}{2}

Decimal Notation

d = 1.54138 \dots, d = - 4.54138 \dots

Solution steps

d^{2} + 3 d - 7 = 0

Solve with the quadratic formula

d_{1, 2} = \frac{- 3 \pm 3 ^{2} - 4 \cdot 1 \cdot ( - 7 )}{2 \cdot 1}

3^{2} - 4 \cdot 1 \cdot (- 7) = 37

d_{1, 2} = \frac{- 3 \pm 37}{2 \cdot 1}

Separate the solutions

d_{1} = \frac{- 3 + 37}{2 \cdot 1}, d_{2} = \frac{- 3 - 37}{2 \cdot 1}

d = \frac{- 3 + 37}{2 \cdot 1} : \frac{- 3 + 37}{2}

d = \frac{- 3 - 37}{2 \cdot 1} : \frac{- 3 - 37}{2}

The solutions to the quadratic equation are:

d = \frac{- 3 + 37}{2}, d = \frac{- 3 - 37}{2}

x, f (x + 1) = - 3 x^{2} - 1

2.6 \cdot 9.8 (3.5 + x) = 0.5 \cdot 512 (x^{2})

(z + 1)^{2} = 2 z^{2}

- 6.5 q^{2} + 390 q - 2000 = 0

x, x^{2} - 3 x y + 2 y^{2} = 0

What are the solutions to the equation d^2+3d-7=0 ?
The solutions to the equation d^2+3d-7=0 are d=(-3+sqrt(37))/2 ,d=(-3-sqrt(37))/2
Find the zeros of d^2+3d-7=0
The zeros of d^2+3d-7=0 are d=(-3+sqrt(37))/2 ,d=(-3-sqrt(37))/2