What are the solutions to the equation 7u^2+2u-15=0 ?

The solutions to the equation 7u^2+2u-15=0 are u=(-1+sqrt(106))/7 ,u=-(1+sqrt(106))/7

Find the zeros of 7u^2+2u-15=0

The zeros of 7u^2+2u-15=0 are u=(-1+sqrt(106))/7 ,u=-(1+sqrt(106))/7

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7 u^{2} + 2 u - 15 = 0

u = \frac{- 1 + 106}{7}, u = - \frac{1 + 106}{7}

Decimal Notation

u = 1.32794 \dots, u = - 1.61366 \dots

Solution steps

7 u^{2} + 2 u - 15 = 0

Solve with the quadratic formula

u_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 \cdot 7 ( - 15 )}{2 \cdot 7}

2^{2} - 4 \cdot 7 (- 15) = 2106

u_{1, 2} = \frac{- 2 \pm 2 106}{2 \cdot 7}

Separate the solutions

u_{1} = \frac{- 2 + 2 106}{2 \cdot 7}, u_{2} = \frac{- 2 - 2 106}{2 \cdot 7}

u = \frac{- 2 + 2 106}{2 \cdot 7} : \frac{- 1 + 106}{7}

u = \frac{- 2 - 2 106}{2 \cdot 7} : - \frac{1 + 106}{7}

The solutions to the quadratic equation are:

u = \frac{- 1 + 106}{7}, u = - \frac{1 + 106}{7}

3 x^{2} - 6 x + 17

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

48 - (10 - 3 + x^{2}) + 2 xx = 4

- x^{2} - 3 x = 4

3 m^{2} - 2 = 1

What are the solutions to the equation 7u^2+2u-15=0 ?
The solutions to the equation 7u^2+2u-15=0 are u=(-1+sqrt(106))/7 ,u=-(1+sqrt(106))/7
Find the zeros of 7u^2+2u-15=0
The zeros of 7u^2+2u-15=0 are u=(-1+sqrt(106))/7 ,u=-(1+sqrt(106))/7