What are the solutions to the equation 5a^2-14a=-280 ?

The solutions to the equation 5a^2-14a=-280 are a= 7/5+i(sqrt(1351))/5 ,a= 7/5-i(sqrt(1351))/5

Find the zeros of 5a^2-14a=-280

The zeros of 5a^2-14a=-280 are a= 7/5+i(sqrt(1351))/5 ,a= 7/5-i(sqrt(1351))/5

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5 a^{2} - 14 a = - 280

a = \frac{7}{5} + i \frac{1351}{5}, a = \frac{7}{5} - i \frac{1351}{5}

Solution steps

5 a^{2} - 14 a = - 280

Move

280

to the left side

5 a^{2} - 14 a + 280 = 0

Solve with the quadratic formula

a_{1, 2} = \frac{- ( - 14 ) \pm ( - 14 ) ^{2} - 4 \cdot 5 \cdot 280}{2 \cdot 5}

Simplify

(- 14)^{2} - 4 \cdot 5 \cdot 280 : 21351 i

a_{1, 2} = \frac{- ( - 14 ) \pm 2 1351 i}{2 \cdot 5}

Separate the solutions

a_{1} = \frac{- ( - 14 ) + 2 1351 i}{2 \cdot 5}, a_{2} = \frac{- ( - 14 ) - 2 1351 i}{2 \cdot 5}

a = \frac{- ( - 14 ) + 2 1351 i}{2 \cdot 5} : \frac{7}{5} + i \frac{1351}{5}

a = \frac{- ( - 14 ) - 2 1351 i}{2 \cdot 5} : \frac{7}{5} - i \frac{1351}{5}

The solutions to the quadratic equation are:

a = \frac{7}{5} + i \frac{1351}{5}, a = \frac{7}{5} - i \frac{1351}{5}

0. 5^{10} < 5^{2}

- 2 - x + 4 (- 2 x - 3) = - 8 x + 1

- \frac{9}{4} + 1

- 26 x^{2} y^{4} z^{5} + 41 x^{4} y z^{3}

What are the solutions to the equation 5a^2-14a=-280 ?
The solutions to the equation 5a^2-14a=-280 are a= 7/5+i(sqrt(1351))/5 ,a= 7/5-i(sqrt(1351))/5
Find the zeros of 5a^2-14a=-280
The zeros of 5a^2-14a=-280 are a= 7/5+i(sqrt(1351))/5 ,a= 7/5-i(sqrt(1351))/5