8c^2+11c+7=6c^2 — step-by-step solution

1) Let's solve in

ℝ

8 c^{2} + 11 c + 7 = 6 c^{2}

8 c^{2}

+ 11 c

+ 7

- 6 c^{2} = 0

ⓘ Like terms are grouped by adding their coefficients:

8 c^{2} - 6 c^{2}

2 c^{2}

2 c^{2} + 11 c + 7 = 0

Given the quadratic form

A c^{2} + B c + C

, let's compute the discriminant:

Δ = B^{2} - 4 A C

. Here

A = 2

B = 11

and

C = 7

Δ = 11^{2} - 4 \cdot 2 \cdot 7

ⓘ Powers are computed before multiplications, additions and subtractions. An integer power is a repeated product:

11^{2}

11 \cdot 11

121

■Multiplications are done before additions and subtractions:

4 \cdot 2

8

Δ = 121 - 8 \cdot 7

ⓘ Multiplications are done before additions and subtractions:

8 \cdot 7

56

Δ = 121 - 56

ⓘ Start from

121

and move back

56

on the number line:

121 - 56

65

Δ = 65

Since

Δ > 0

, the equation has two roots:

The quadratic formula gives

c_{1}

\frac{- B - \sqrt{Δ}}{2 A}

\frac{- 11 - \sqrt{65}}{2 \cdot 2}

c_{1} = \frac{- 11 - \sqrt{65}}{2 \cdot 2}

ⓘ The minus sign is factored out, changing the sign of each term:

- 11 - \sqrt{65}

- (11 + \sqrt{65})

ⅰ

2 \cdot 2

4

c_{1} = - \frac{11 + \sqrt{65}}{4}

The quadratic formula gives

c_{2}

\frac{- B + \sqrt{Δ}}{2 A}

\frac{- 11 + \sqrt{65}}{2 \cdot 2}

c_{2} = \frac{- 11 + \sqrt{65}}{2 \cdot 2}

ⅰ

2 \cdot 2

4

c_{2} = \frac{- 11 + \sqrt{65}}{4}

▶Solutions obtained:

c = {- \frac{11 + \sqrt{65}}{4}, \frac{\sqrt{65} - 11}{4}}

2) The numeric value of:

c = {- \frac{11 + \sqrt{65}}{4}, \frac{\sqrt{65} - 11}{4}}

is:

c = {\approx - \frac{11 + 8.0623}{4}, \approx \frac{8.0623 - 11}{4}}

c = {\approx - \frac{19.0623}{4}, \approx - \frac{2.9377}{4}}

▷

c = {\approx -4.7656, \approx -0.7344}