What are the solutions to the equation 4x^2+23x=-28 ?

The solutions to the equation 4x^2+23x=-28 are x=-7/4 ,x=-4

Find the zeros of 4x^2+23x=-28

The zeros of 4x^2+23x=-28 are x=-7/4 ,x=-4

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4 x^{2} + 23 x = - 28

x = - \frac{7}{4}, x = - 4

Decimal

x = - 1.75, x = - 4

Solution steps

4 x^{2} + 23 x = - 28

Move

28

to the left side

4 x^{2} + 23 x + 28 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 23 \pm 2 3 ^{2} - 4 \cdot 4 \cdot 28}{2 \cdot 4}

2 3^{2} - 4 \cdot 4 \cdot 28 = 9

x_{1, 2} = \frac{- 23 \pm 9}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- 23 + 9}{2 \cdot 4}, x_{2} = \frac{- 23 - 9}{2 \cdot 4}

x = \frac{- 23 + 9}{2 \cdot 4} : - \frac{7}{4}

x = \frac{- 23 - 9}{2 \cdot 4} : - 4

The solutions to the quadratic equation are:

x = - \frac{7}{4}, x = - 4

2 x^{2} + 3 x + 0.5 = 0

2 (x - 7)^{2} - 31 = 11

0 = 50 + 20 \cdot t - 4.9 \cdot t^{2}

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

- \frac{1}{2} (x - 7)^{2} = 96

What are the solutions to the equation 4x^2+23x=-28 ?
The solutions to the equation 4x^2+23x=-28 are x=-7/4 ,x=-4
Find the zeros of 4x^2+23x=-28
The zeros of 4x^2+23x=-28 are x=-7/4 ,x=-4