What are the solutions to the equation 2c^2-5c-14=-5 ?

The solutions to the equation 2c^2-5c-14=-5 are c=(5+sqrt(97))/4 ,c=(5-sqrt(97))/4

Find the zeros of 2c^2-5c-14=-5

The zeros of 2c^2-5c-14=-5 are c=(5+sqrt(97))/4 ,c=(5-sqrt(97))/4

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2 c^{2} - 5 c - 14 = - 5

c = \frac{5 + 97}{4}, c = \frac{5 - 97}{4}

Decimal Notation

c = 3.71221 \dots, c = - 1.21221 \dots

Solution steps

2 c^{2} - 5 c - 14 = - 5

Move

5

to the left side

2 c^{2} - 5 c - 9 = 0

Solve with the quadratic formula

c_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 \cdot 2 ( - 9 )}{2 \cdot 2}

(- 5)^{2} - 4 \cdot 2 (- 9) = 97

c_{1, 2} = \frac{- ( - 5 ) \pm 97}{2 \cdot 2}

Separate the solutions

c_{1} = \frac{- ( - 5 ) + 97}{2 \cdot 2}, c_{2} = \frac{- ( - 5 ) - 97}{2 \cdot 2}

c = \frac{- ( - 5 ) + 97}{2 \cdot 2} : \frac{5 + 97}{4}

c = \frac{- ( - 5 ) - 97}{2 \cdot 2} : \frac{5 - 97}{4}

The solutions to the quadratic equation are:

c = \frac{5 + 97}{4}, c = \frac{5 - 97}{4}

x^{2} - 6 x + 8 = 0

5 r^{2} + r - 2 = 0

- 7 x^{2} - 10 = - 12 x

x, 7 x^{2} - 5 x y + y^{3} = 3

x^{2} + 10 x + 29

What are the solutions to the equation 2c^2-5c-14=-5 ?
The solutions to the equation 2c^2-5c-14=-5 are c=(5+sqrt(97))/4 ,c=(5-sqrt(97))/4
Find the zeros of 2c^2-5c-14=-5
The zeros of 2c^2-5c-14=-5 are c=(5+sqrt(97))/4 ,c=(5-sqrt(97))/4