What are the solutions to the equation 5r^2-5r=-8 ?

The solutions to the equation 5r^2-5r=-8 are r= 1/2+i(3sqrt(15))/(10),r= 1/2-i(3sqrt(15))/(10)

Find the zeros of 5r^2-5r=-8

The zeros of 5r^2-5r=-8 are r= 1/2+i(3sqrt(15))/(10),r= 1/2-i(3sqrt(15))/(10)

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5 r^{2} - 5 r = - 8

r = \frac{1}{2} + i \frac{3 15}{10}, r = \frac{1}{2} - i \frac{3 15}{10}

Solution steps

5 r^{2} - 5 r = - 8

Move

8

to the left side

5 r^{2} - 5 r + 8 = 0

Solve with the quadratic formula

r_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 \cdot 5 \cdot 8}{2 \cdot 5}

Simplify

(- 5)^{2} - 4 \cdot 5 \cdot 8 : 315 i

r_{1, 2} = \frac{- ( - 5 ) \pm 3 15 i}{2 \cdot 5}

Separate the solutions

r_{1} = \frac{- ( - 5 ) + 3 15 i}{2 \cdot 5}, r_{2} = \frac{- ( - 5 ) - 3 15 i}{2 \cdot 5}

r = \frac{- ( - 5 ) + 3 15 i}{2 \cdot 5} : \frac{1}{2} + i \frac{3 15}{10}

r = \frac{- ( - 5 ) - 3 15 i}{2 \cdot 5} : \frac{1}{2} - i \frac{3 15}{10}

The solutions to the quadratic equation are:

r = \frac{1}{2} + i \frac{3 15}{10}, r = \frac{1}{2} - i \frac{3 15}{10}

x, x^{2} y^{2} = - 1

x^{2} - 14 = 2 x

8 t^{2} - 2 t = 3

3 (x + 1)^{2} = 81

8 x^{2} + 16 x - 8 = 0

What are the solutions to the equation 5r^2-5r=-8 ?
The solutions to the equation 5r^2-5r=-8 are r= 1/2+i(3sqrt(15))/(10),r= 1/2-i(3sqrt(15))/(10)
Find the zeros of 5r^2-5r=-8
The zeros of 5r^2-5r=-8 are r= 1/2+i(3sqrt(15))/(10),r= 1/2-i(3sqrt(15))/(10)