What are the solutions to the equation (2x+1)^2-x^2+5x-10=0 ?

The solutions to the equation (2x+1)^2-x^2+5x-10=0 are x=(-3+sqrt(21))/2 ,x=-(3+sqrt(21))/2

Find the zeros of (2x+1)^2-x^2+5x-10=0

The zeros of (2x+1)^2-x^2+5x-10=0 are x=(-3+sqrt(21))/2 ,x=-(3+sqrt(21))/2

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(2 x + 1)^{2} - x^{2} + 5 x - 10 = 0

x = \frac{- 3 + 21}{2}, x = - \frac{3 + 21}{2}

Solution steps

(2 x + 1)^{2} - x^{2} + 5 x - 10 = 0

Expand

(2 x + 1)^{2} - x^{2} + 5 x - 10 : 3 x^{2} + 9 x - 9

3 x^{2} + 9 x - 9 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 9 \pm 9 ^{2} - 4 \cdot 3 ( - 9 )}{2 \cdot 3}

9^{2} - 4 \cdot 3 (- 9) = 321

x_{1, 2} = \frac{- 9 \pm 3 21}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- 9 + 3 21}{2 \cdot 3}, x_{2} = \frac{- 9 - 3 21}{2 \cdot 3}

x = \frac{- 9 + 3 21}{2 \cdot 3} : \frac{- 3 + 21}{2}

x = \frac{- 9 - 3 21}{2 \cdot 3} : - \frac{3 + 21}{2}

The solutions to the quadratic equation are:

x = \frac{- 3 + 21}{2}, x = - \frac{3 + 21}{2}

64 a z^{2} - 81 a^{3} = 0

- 0.005 x^{2} + 0.101 x - 0.193 = 0

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

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

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

What are the solutions to the equation (2x+1)^2-x^2+5x-10=0 ?
The solutions to the equation (2x+1)^2-x^2+5x-10=0 are x=(-3+sqrt(21))/2 ,x=-(3+sqrt(21))/2
Find the zeros of (2x+1)^2-x^2+5x-10=0
The zeros of (2x+1)^2-x^2+5x-10=0 are x=(-3+sqrt(21))/2 ,x=-(3+sqrt(21))/2