What are the solutions to the equation -4a^2-28a+40=0 ?

The solutions to the equation -4a^2-28a+40=0 are a=-(7+sqrt(89))/2 ,a=(sqrt(89)-7)/2

Find the zeros of -4a^2-28a+40=0

The zeros of -4a^2-28a+40=0 are a=-(7+sqrt(89))/2 ,a=(sqrt(89)-7)/2

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- 4 a^{2} - 28 a + 40 = 0

a = - \frac{7 + 89}{2}, a = \frac{89 - 7}{2}

Decimal Notation

a = - 8.21699 \dots, a = 1.21699 \dots

Solution steps

- 4 a^{2} - 28 a + 40 = 0

Solve with the quadratic formula

a_{1, 2} = \frac{- ( - 28 ) \pm ( - 28 ) ^{2} - 4 ( - 4 ) \cdot 40}{2 ( - 4 )}

(- 28)^{2} - 4 (- 4) \cdot 40 = 489

a_{1, 2} = \frac{- ( - 28 ) \pm 4 89}{2 ( - 4 )}

Separate the solutions

a_{1} = \frac{- ( - 28 ) + 4 89}{2 ( - 4 )}, a_{2} = \frac{- ( - 28 ) - 4 89}{2 ( - 4 )}

a = \frac{- ( - 28 ) + 4 89}{2 ( - 4 )} : - \frac{7 + 89}{2}

a = \frac{- ( - 28 ) - 4 89}{2 ( - 4 )} : \frac{89 - 7}{2}

The solutions to the quadratic equation are:

a = - \frac{7 + 89}{2}, a = \frac{89 - 7}{2}

\frac{5 ^{8}}{5 ^{4}}

x + 1 = 9

lo g_{1.1} (2) = x

e^{0.1}

(- \frac{2}{5} - \frac{8}{7}) \frac{1}{3} + (\frac{1}{5} - \frac{5}{7}) \frac{1}{6}

What are the solutions to the equation -4a^2-28a+40=0 ?
The solutions to the equation -4a^2-28a+40=0 are a=-(7+sqrt(89))/2 ,a=(sqrt(89)-7)/2
Find the zeros of -4a^2-28a+40=0
The zeros of -4a^2-28a+40=0 are a=-(7+sqrt(89))/2 ,a=(sqrt(89)-7)/2