What are the solutions to the equation 4c^2-2c=23 ?

The solutions to the equation 4c^2-2c=23 are c=(1+sqrt(93))/4 ,c=(1-sqrt(93))/4

Find the zeros of 4c^2-2c=23

The zeros of 4c^2-2c=23 are c=(1+sqrt(93))/4 ,c=(1-sqrt(93))/4

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4 c^{2} - 2 c = 23

c = \frac{1 + 93}{4}, c = \frac{1 - 93}{4}

Decimal

c = 2.66091 \dots, c = - 2.16091 \dots

Solution steps

4 c^{2} - 2 c = 23

Move

23

to the left side

4 c^{2} - 2 c - 23 = 0

Solve with the quadratic formula

c_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 \cdot 4 ( - 23 )}{2 \cdot 4}

(- 2)^{2} - 4 \cdot 4 (- 23) = 293

c_{1, 2} = \frac{- ( - 2 ) \pm 2 93}{2 \cdot 4}

Separate the solutions

c_{1} = \frac{- ( - 2 ) + 2 93}{2 \cdot 4}, c_{2} = \frac{- ( - 2 ) - 2 93}{2 \cdot 4}

c = \frac{- ( - 2 ) + 2 93}{2 \cdot 4} : \frac{1 + 93}{4}

c = \frac{- ( - 2 ) - 2 93}{2 \cdot 4} : \frac{1 - 93}{4}

The solutions to the quadratic equation are:

c = \frac{1 + 93}{4}, c = \frac{1 - 93}{4}

2 x^{2} + 7 x - 4 = 0

2 x^{2} + 5 x - 168 = 0

x^{2} + 27 = 31

d^{2} = 25

y^{2} + 8 = 0

What are the solutions to the equation 4c^2-2c=23 ?
The solutions to the equation 4c^2-2c=23 are c=(1+sqrt(93))/4 ,c=(1-sqrt(93))/4
Find the zeros of 4c^2-2c=23
The zeros of 4c^2-2c=23 are c=(1+sqrt(93))/4 ,c=(1-sqrt(93))/4