Question

lets simplify.
$8x^2 + 14x + 5 = 0$
$x = \frac{\square\vert\square \pm \square}{\square}$

Explanation:

Step1: Identify coefficients

For quadratic equation \(ax^2 + bx + c = 0\), here \(a = 8\), \(b = 14\), \(c = 5\).

Step2: Use quadratic formula

Quadratic formula is \(x=\frac{-b\pm\sqrt{b^2 - 4ac}}{2a}\). First calculate discriminant \(D = b^2-4ac=(14)^2 - 4\times8\times5=196 - 160 = 36\). Then \(\sqrt{D}=\sqrt{36}=6\). Now substitute into formula: \(x=\frac{- 14\pm6}{2\times8}\).

Answer:

\(x=\frac{-14\pm6}{16}\) (So the boxes from left to right, top to bottom: \(-14\), (the middle top box is part of the numerator's first term, already included in \(-14\) structure), \(6\), \(16\))