What are the solutions to the equation 4x^2-28x+53=0 ?

The solutions to the equation 4x^2-28x+53=0 are x= 7/2+i,x= 7/2-i

Find the zeros of 4x^2-28x+53=0

The zeros of 4x^2-28x+53=0 are x= 7/2+i,x= 7/2-i

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

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

Solution steps

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

Solve with the quadratic formula

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

Simplify

(- 28)^{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

x^{2} + 4 x - 2

x^{2} + 5 + 6 = 0

s^{2} + 2 s + 3

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

5 x^{2} = 17 x - 14

What are the solutions to the equation 4x^2-28x+53=0 ?
The solutions to the equation 4x^2-28x+53=0 are x= 7/2+i,x= 7/2-i
Find the zeros of 4x^2-28x+53=0
The zeros of 4x^2-28x+53=0 are x= 7/2+i,x= 7/2-i