What are the solutions to the equation (5x^2-48x)/4+34=0 ?

The solutions to the equation (5x^2-48x)/4+34=0 are x= 24/5+i(2sqrt(26))/5 ,x= 24/5-i(2sqrt(26))/5

Find the zeros of (5x^2-48x)/4+34=0

The zeros of (5x^2-48x)/4+34=0 are x= 24/5+i(2sqrt(26))/5 ,x= 24/5-i(2sqrt(26))/5

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\frac{5 x ^{2} - 48 x}{4} + 34 = 0

x = \frac{24}{5} + i \frac{2 26}{5}, x = \frac{24}{5} - i \frac{2 26}{5}

Solution steps

\frac{5 x ^{2} - 48 x}{4} + 34 = 0

Multiply both sides by

4

5 x^{2} - 48 x + 136 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 48 ) \pm ( - 48 ) ^{2} - 4 \cdot 5 \cdot 136}{2 \cdot 5}

Simplify

(- 48)^{2} - 4 \cdot 5 \cdot 136 : 426 i

x_{1, 2} = \frac{- ( - 48 ) \pm 4 26 i}{2 \cdot 5}

Separate the solutions

x_{1} = \frac{- ( - 48 ) + 4 26 i}{2 \cdot 5}, x_{2} = \frac{- ( - 48 ) - 4 26 i}{2 \cdot 5}

x = \frac{- ( - 48 ) + 4 26 i}{2 \cdot 5} : \frac{24}{5} + i \frac{2 26}{5}

x = \frac{- ( - 48 ) - 4 26 i}{2 \cdot 5} : \frac{24}{5} - i \frac{2 26}{5}

The solutions to the quadratic equation are:

x = \frac{24}{5} + i \frac{2 26}{5}, x = \frac{24}{5} - i \frac{2 26}{5}

- 4 (- 2)^{2}

\frac{1}{3} + 2

z^{2} + 6 z = - 8

112 = 112 + 96 t - 16 t^{2}

12 x^{3} - 9 x^{2}

What are the solutions to the equation (5x^2-48x)/4+34=0 ?
The solutions to the equation (5x^2-48x)/4+34=0 are x= 24/5+i(2sqrt(26))/5 ,x= 24/5-i(2sqrt(26))/5
Find the zeros of (5x^2-48x)/4+34=0
The zeros of (5x^2-48x)/4+34=0 are x= 24/5+i(2sqrt(26))/5 ,x= 24/5-i(2sqrt(26))/5