Solve (4+2+3)x+(4+2+3x)x=100

Question

Solve the quadratic equation

Solve using the quadratic formula

Solve by completing the square

Solve using the PQ formula

x_{1} = - \frac{15 + 5 57}{6}, x_{2} = \frac{- 15 + 5 57}{6}

Alternative Form

x_{1} \approx - 8.791529, x_{2} \approx 3.791529

Evaluate

(4 + 2 + 3) x + (4 + 2 + 3 x) x = 100

Simplify

More Steps

9 x + x (6 + 3 x) = 100

Expand the expression

More Steps

15 x + 3 x^{2} = 100

Move the expression to the left side

15 x + 3 x^{2} - 100 = 0

Rewrite in standard form

3 x^{2} + 15 x - 100 = 0

Substitute a= 3,b= 15 and c= - 100 into the quadratic formula x = \frac{- b \pm b ^{2} - 4 a c}{2 a}

x = \frac{- 15 \pm 1 5 ^{2} - 4 \times 3 ( - 100 )}{2 \times 3}

Simplify the expression

x = \frac{- 15 \pm 1 5 ^{2} - 4 \times 3 ( - 100 )}{6}

Simplify the expression

More Steps

x = \frac{- 15 \pm 1425}{6}

Simplify the radical expression

More Steps

x = \frac{- 15 \pm 5 57}{6}

Separate the equation into 2 possible cases

x = \frac{- 15 + 5 57}{6} x = \frac{- 15 - 5 57}{6}

Use \frac{- a}{b} = \frac{a}{- b} = - \frac{a}{b} to rewrite the fraction

x = \frac{- 15 + 5 57}{6} x = - \frac{15 + 5 57}{6}

Solution

x_{1} = - \frac{15 + 5 57}{6}, x_{2} = \frac{- 15 + 5 57}{6}

Alternative Form

x_{1} \approx - 8.791529, x_{2} \approx 3.791529

Show Solution