What are the solutions to the equation 4x^2+10x-5=x^2-x-15 ?

The solutions to the equation 4x^2+10x-5=x^2-x-15 are x=-5/3 ,x=-2

Find the zeros of 4x^2+10x-5=x^2-x-15

The zeros of 4x^2+10x-5=x^2-x-15 are x=-5/3 ,x=-2

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4 x^{2} + 10 x - 5 = x^{2} - x - 15

x = - \frac{5}{3}, x = - 2

Solution steps

4 x^{2} + 10 x - 5 = x^{2} - x - 15

Move

15

to the left side

4 x^{2} + 10 x + 10 = x^{2} - x

Move

x

to the left side

4 x^{2} + 11 x + 10 = x^{2}

Move

x^{2}

to the left side

3 x^{2} + 11 x + 10 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 11 \pm 1 1 ^{2} - 4 \cdot 3 \cdot 10}{2 \cdot 3}

1 1^{2} - 4 \cdot 3 \cdot 10 = 1

x_{1, 2} = \frac{- 11 \pm 1}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- 11 + 1}{2 \cdot 3}, x_{2} = \frac{- 11 - 1}{2 \cdot 3}

x = \frac{- 11 + 1}{2 \cdot 3} : - \frac{5}{3}

x = \frac{- 11 - 1}{2 \cdot 3} : - 2

The solutions to the quadratic equation are:

x = - \frac{5}{3}, x = - 2

x^{2} - \frac{49}{4}

\frac{2}{3} + \frac{1}{2}

v^{2} + 9 v + 18

100 f - 100 = 50 f + 100

- 8 < n - 3 < 3

What are the solutions to the equation 4x^2+10x-5=x^2-x-15 ?
The solutions to the equation 4x^2+10x-5=x^2-x-15 are x=-5/3 ,x=-2
Find the zeros of 4x^2+10x-5=x^2-x-15
The zeros of 4x^2+10x-5=x^2-x-15 are x=-5/3 ,x=-2