What are the solutions to the equation 3x^2+24x+53=5 ?

The solutions to the equation 3x^2+24x+53=5 are x=-4

Find the zeros of 3x^2+24x+53=5

The zeros of 3x^2+24x+53=5 are x=-4

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3 x^{2} + 24 x + 53 = 5

x = - 4

Solution steps

3 x^{2} + 24 x + 53 = 5

Move

5

to the left side

3 x^{2} + 24 x + 48 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 24 \pm 2 4 ^{2} - 4 \cdot 3 \cdot 48}{2 \cdot 3}

2 4^{2} - 4 \cdot 3 \cdot 48 = 0

x_{1, 2} = \frac{- 24 \pm 0}{2 \cdot 3}

x = \frac{- 24}{2 \cdot 3}

\frac{- 24}{2 \cdot 3} = - 4

x = - 4

The solution to the quadratic equation is:

x = - 4

21 a - 7 a^{2}

\frac{5 x y ^{8}}{5 x ^{6} y ^{4}}

\frac{3}{x - 1} - \frac{4}{x} \geq 1

6^{2} + 8^{2} = c^{2}

3 x^{3} - 5 x^{2} - 4 x + 4

What are the solutions to the equation 3x^2+24x+53=5 ?
The solutions to the equation 3x^2+24x+53=5 are x=-4
Find the zeros of 3x^2+24x+53=5
The zeros of 3x^2+24x+53=5 are x=-4