What are the solutions to the equation 0=x^2+3x-4 ?

The solutions to the equation 0=x^2+3x-4 are x=1,x=-4

Find the zeros of 0=x^2+3x-4

The zeros of 0=x^2+3x-4 are x=1,x=-4

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0 = x^{2} + 3 x - 4

x = 1, x = - 4

Solution steps

0 = x^{2} + 3 x - 4

Switch sides

x^{2} + 3 x - 4 = 0

Solve with the quadratic formula

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

3^{2} - 4 \cdot 1 \cdot (- 4) = 5

x_{1, 2} = \frac{- 3 \pm 5}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 3 + 5}{2 \cdot 1}, x_{2} = \frac{- 3 - 5}{2 \cdot 1}

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

x = \frac{- 3 - 5}{2 \cdot 1} : - 4

The solutions to the quadratic equation are:

x = 1, x = - 4

4 x^{2} + 20 x + 13 = 0

2 x^{2} + x \geq 6

6 - x > 0

- 12 - 4

(\frac{3 m ^{5} r ^{3}}{4 p ^{8}})^{4}

What are the solutions to the equation 0=x^2+3x-4 ?
The solutions to the equation 0=x^2+3x-4 are x=1,x=-4
Find the zeros of 0=x^2+3x-4
The zeros of 0=x^2+3x-4 are x=1,x=-4