What are the solutions to the equation 4x^2-48x+335=7 ?

The solutions to the equation 4x^2-48x+335=7 are x=6+sqrt(46)i,x=6-sqrt(46)i

Find the zeros of 4x^2-48x+335=7

The zeros of 4x^2-48x+335=7 are x=6+sqrt(46)i,x=6-sqrt(46)i

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4 x^{2} - 48 x + 335 = 7

x = 6 + 46 i, x = 6 - 46 i

Solution steps

4 x^{2} - 48 x + 335 = 7

Move

7

to the left side

4 x^{2} - 48 x + 328 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 48 ) \pm ( - 48 ) ^{2} - 4 \cdot 4 \cdot 328}{2 \cdot 4}

Simplify

(- 48)^{2} - 4 \cdot 4 \cdot 328 : 846 i

x_{1, 2} = \frac{- ( - 48 ) \pm 8 46 i}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- ( - 48 ) + 8 46 i}{2 \cdot 4}, x_{2} = \frac{- ( - 48 ) - 8 46 i}{2 \cdot 4}

x = \frac{- ( - 48 ) + 8 46 i}{2 \cdot 4} : 6 + 46 i

x = \frac{- ( - 48 ) - 8 46 i}{2 \cdot 4} : 6 - 46 i

The solutions to the quadratic equation are:

x = 6 + 46 i, x = 6 - 46 i

260 + 13 x - 13 x^{2} = 13 x^{2} - 65 x

- \frac{1}{2} (x + 1)^{2} + 8 = - 1

x^{2} + 2 x + 63 = 0

4 x^{2} + 20 x = 2

0 = 3 x^{2} - 36 x + 81

What are the solutions to the equation 4x^2-48x+335=7 ?
The solutions to the equation 4x^2-48x+335=7 are x=6+sqrt(46)i,x=6-sqrt(46)i
Find the zeros of 4x^2-48x+335=7
The zeros of 4x^2-48x+335=7 are x=6+sqrt(46)i,x=6-sqrt(46)i