A triangular-shaped wall has a base of 2x + 4 and a height of x + 3. The area of the triangle is 56 in.2. What is the value of x?
Area of the triangle = b*h/2 = (2x+4)(x+3)/2 = 2x^2+10x+12 all divided by 2 = x^2+5x+6.
\(x^2+5x+6 = 56, x^2+5x-50 = 0, (x-5)(x+10) = 0\)
can you solve from here?
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