# What are the real solution of equation? 2x^2+6x+11=6 ?

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### 2 Answers

To find the real solution of equation? 2x^2+6x+11=6 .

We rewrite the equation by subtracting 6 from both sides.

2x^2+6x+11-6 = 0

2x^2+6x+5 = 0

The discriminant of the equation is (coefficient of x)^2 - 4*(coefficient of x^2)*(constant term) = 6^2-4*2*5 = 36 - 40 = -4 which is negative.

Since the discriminant is negative both roots are not real.

Therefore the the quadratic equation has no real solution as its both roots are not real.

First, we'll move all terms to the left side:

2x^2 + 6x + 11 - 6 = 0

We'll combine like terms:

2x^2 + 6x + 5 = 0

Now, we'll verify if the equation has real solutions. For this reason, we'll calculate the discriminant of the equation.

delta = b^2 - 4ac, where ab,c, are the coefficients of the equation:

ax^2 + bx + c = 0

We'll identify a,b,c:

a = 2

b = 6

c = 5

delta = 36 - 40 = -4 < 0

**Since delta is negative, then the equation has no real roots, but it has complex roots.**