What are the solutions to the equation 4x^2+28x=-53 ?

The solutions to the equation 4x^2+28x=-53 are x=-7/2+i,x=-7/2-i

Find the zeros of 4x^2+28x=-53

The zeros of 4x^2+28x=-53 are x=-7/2+i,x=-7/2-i

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4 x^{2} + 28 x = - 53

x = - \frac{7}{2} + i, x = - \frac{7}{2} - i

Solution steps

4 x^{2} + 28 x = - 53

Move

53

to the left side

4 x^{2} + 28 x + 53 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 28 \pm 2 8 ^{2} - 4 \cdot 4 \cdot 53}{2 \cdot 4}

Simplify

2 8^{2} - 4 \cdot 4 \cdot 53 : 8 i

x_{1, 2} = \frac{- 28 \pm 8 i}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- 28 + 8 i}{2 \cdot 4}, x_{2} = \frac{- 28 - 8 i}{2 \cdot 4}

x = \frac{- 28 + 8 i}{2 \cdot 4} : - \frac{7}{2} + i

x = \frac{- 28 - 8 i}{2 \cdot 4} : - \frac{7}{2} - i

The solutions to the quadratic equation are:

x = - \frac{7}{2} + i, x = - \frac{7}{2} - i

- 5 x (x + 6) - 10 = 10

6 x^{2} = 16 x - 8

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

0.36 x^{2} + 10.236 x + 18.2 = 0

- 2 x^{2} = - 10

What are the solutions to the equation 4x^2+28x=-53 ?
The solutions to the equation 4x^2+28x=-53 are x=-7/2+i,x=-7/2-i
Find the zeros of 4x^2+28x=-53
The zeros of 4x^2+28x=-53 are x=-7/2+i,x=-7/2-i