What are the solutions to the equation 18x^2+48x=-32 ?

The solutions to the equation 18x^2+48x=-32 are x=-4/3

Find the zeros of 18x^2+48x=-32

The zeros of 18x^2+48x=-32 are x=-4/3

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18 x^{2} + 48 x = - 32

x = - \frac{4}{3}

Decimal Notation

x = - 1.33333 \dots

Solution steps

18 x^{2} + 48 x = - 32

Move

32

to the left side

18 x^{2} + 48 x + 32 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 48 \pm 4 8 ^{2} - 4 \cdot 18 \cdot 32}{2 \cdot 18}

4 8^{2} - 4 \cdot 18 \cdot 32 = 0

x_{1, 2} = \frac{- 48 \pm 0}{2 \cdot 18}

x = \frac{- 48}{2 \cdot 18}

\frac{- 48}{2 \cdot 18} = - \frac{4}{3}

x = - \frac{4}{3}

The solution to the quadratic equation is:

x = - \frac{4}{3}

m, 2 (m^{2} + n^{2}) = 58

- \frac{1}{4} b^{2} - b^{2} = - 1

6 x - 8 = 2 x^{2}

6 x - 12 x^{2} = 0

3 q^{2} - 300 q - 112500 = 0

What are the solutions to the equation 18x^2+48x=-32 ?
The solutions to the equation 18x^2+48x=-32 are x=-4/3
Find the zeros of 18x^2+48x=-32
The zeros of 18x^2+48x=-32 are x=-4/3