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

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

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

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

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4 x^{2} + 53 = 28 x

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

Solution steps

4 x^{2} + 53 = 28 x

Move

28 x

to the left side

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

Write in the standard form

a x^{2} + b x + c = 0

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

9 z^{2} - 27 z + 20 = 0

y, x^{2} + y^{2} - x y = 1

y, x^{2} + 2 x y - y^{2} + x = 36

\frac{x ^{2}}{2} - 4 x = 3 + \frac{x ^{2} - 12}{4}

x^{2} - 100 = 0

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