What are the solutions to the equation 4x^2+19x=-12 ?

The solutions to the equation 4x^2+19x=-12 are x=-3/4 ,x=-4

Find the zeros of 4x^2+19x=-12

The zeros of 4x^2+19x=-12 are x=-3/4 ,x=-4

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4 x^{2} + 19 x = - 12

x = - \frac{3}{4}, x = - 4

Decimal Notation

x = - 0.75, x = - 4

Solution steps

4 x^{2} + 19 x = - 12

Move

12

to the left side

4 x^{2} + 19 x + 12 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 19 \pm 1 9 ^{2} - 4 \cdot 4 \cdot 12}{2 \cdot 4}

1 9^{2} - 4 \cdot 4 \cdot 12 = 13

x_{1, 2} = \frac{- 19 \pm 13}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- 19 + 13}{2 \cdot 4}, x_{2} = \frac{- 19 - 13}{2 \cdot 4}

x = \frac{- 19 + 13}{2 \cdot 4} : - \frac{3}{4}

x = \frac{- 19 - 13}{2 \cdot 4} : - 4

The solutions to the quadratic equation are:

x = - \frac{3}{4}, x = - 4

x^{2} - 19 x - 3 = 0

10 v^{2} - 20 v + 10 = 0

30 x^{2} - 10 = 5 x^{2} - 1

m^{2} - 2 m - 2 = 0

x, x^{2} + y^{2} - 2 x - 2 y + 2 = 0

What are the solutions to the equation 4x^2+19x=-12 ?
The solutions to the equation 4x^2+19x=-12 are x=-3/4 ,x=-4
Find the zeros of 4x^2+19x=-12
The zeros of 4x^2+19x=-12 are x=-3/4 ,x=-4