What are the solutions to the equation 4z^2+29=24z ?

The solutions to the equation 4z^2+29=24z are z=(6+sqrt(7))/2 ,z=(6-sqrt(7))/2

Find the zeros of 4z^2+29=24z

The zeros of 4z^2+29=24z are z=(6+sqrt(7))/2 ,z=(6-sqrt(7))/2

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4 z^{2} + 29 = 24 z

z = \frac{6 + 7}{2}, z = \frac{6 - 7}{2}

Decimal Notation

z = 4.32287 \dots, z = 1.67712 \dots

Solution steps

4 z^{2} + 29 = 24 z

Move

24 z

to the left side

4 z^{2} + 29 - 24 z = 0

Write in the standard form

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

4 z^{2} - 24 z + 29 = 0

Solve with the quadratic formula

z_{1, 2} = \frac{- ( - 24 ) \pm ( - 24 ) ^{2} - 4 \cdot 4 \cdot 29}{2 \cdot 4}

(- 24)^{2} - 4 \cdot 4 \cdot 29 = 47

z_{1, 2} = \frac{- ( - 24 ) \pm 4 7}{2 \cdot 4}

Separate the solutions

z_{1} = \frac{- ( - 24 ) + 4 7}{2 \cdot 4}, z_{2} = \frac{- ( - 24 ) - 4 7}{2 \cdot 4}

z = \frac{- ( - 24 ) + 4 7}{2 \cdot 4} : \frac{6 + 7}{2}

z = \frac{- ( - 24 ) - 4 7}{2 \cdot 4} : \frac{6 - 7}{2}

The solutions to the quadratic equation are:

z = \frac{6 + 7}{2}, z = \frac{6 - 7}{2}

x^{2} + 6 + 9 = 0

(\frac{2 x}{3 ( x ^{2} ) ^{\frac{2}{3}}})^{2}

2 j^{2} + 9 j + 7

x - 7.6 (0.9 x + 1) \leq 5.6 (1 + 1.1 x)

\frac{1}{x - 4} \geq \frac{3}{x + 1}

What are the solutions to the equation 4z^2+29=24z ?
The solutions to the equation 4z^2+29=24z are z=(6+sqrt(7))/2 ,z=(6-sqrt(7))/2
Find the zeros of 4z^2+29=24z
The zeros of 4z^2+29=24z are z=(6+sqrt(7))/2 ,z=(6-sqrt(7))/2