What are the solutions to the equation a(-2a+14)-6(a^2-12)+2(-7a+12)=0 ?

The solutions to the equation a(-2a+14)-6(a^2-12)+2(-7a+12)=0 are a=-2sqrt(3),a=2sqrt(3)

Find the zeros of a(-2a+14)-6(a^2-12)+2(-7a+12)=0

The zeros of a(-2a+14)-6(a^2-12)+2(-7a+12)=0 are a=-2sqrt(3),a=2sqrt(3)

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a (- 2 a + 14) - 6 (a^{2} - 12) + 2 (- 7 a + 12) = 0

a = - 23, a = 23

Decimal Notation

a = - 3.46410 \dots, a = 3.46410 \dots

Solution steps

a (- 2 a + 14) - 6 (a^{2} - 12) + 2 (- 7 a + 12) = 0

Expand

a (- 2 a + 14) - 6 (a^{2} - 12) + 2 (- 7 a + 12) : - 8 a^{2} + 96

- 8 a^{2} + 96 = 0

Solve with the quadratic formula

a_{1, 2} = \frac{- 0 \pm 0 ^{2} - 4 ( - 8 ) \cdot 96}{2 ( - 8 )}

0^{2} - 4 (- 8) \cdot 96 = 323

a_{1, 2} = \frac{- 0 \pm 32 3}{2 ( - 8 )}

Separate the solutions

a_{1} = \frac{- 0 + 32 3}{2 ( - 8 )}, a_{2} = \frac{- 0 - 32 3}{2 ( - 8 )}

a = \frac{- 0 + 32 3}{2 ( - 8 )} : - 23

a = \frac{- 0 - 32 3}{2 ( - 8 )} : 23

The solutions to the quadratic equation are:

a = - 23, a = 23

\frac{368.2 - 257}{9 - 4}

8 m^{2} + 6 m - 5

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

\frac{5}{2} (2)^{- 2}

- 15 + r = 5

What are the solutions to the equation a(-2a+14)-6(a^2-12)+2(-7a+12)=0 ?
The solutions to the equation a(-2a+14)-6(a^2-12)+2(-7a+12)=0 are a=-2sqrt(3),a=2sqrt(3)
Find the zeros of a(-2a+14)-6(a^2-12)+2(-7a+12)=0
The zeros of a(-2a+14)-6(a^2-12)+2(-7a+12)=0 are a=-2sqrt(3),a=2sqrt(3)