Solve -4.7714x^2 - 0.0884x + 10.014 = 0 for x.

Step-by-step solution

1. 1

   Identify the coefficients. Setting $y=0$ gives a quadratic in standard form with

   $a=-4.7714,\qquad b=-0.0884,\qquad c=10.014$

   Because $a<0$ the parabola opens downward, and since $c>0$ the graph is already above the axis at $x=0$ — so two real roots straddling the origin are expected.

2. 2

   Compute the discriminant.

   $\Delta=b^2-4ac=(-0.0884)^2-4(-4.7714)(10.014)=0.007815+191.123=191.13101$

   It is positive, confirming two distinct real roots. Note how small $b^2$ is compared with $-4ac$; the linear term barely influences the result.

3. 3

   Apply the quadratic formula.

   $x=\frac{-b\pm\sqrt{\Delta}}{2a}=\frac{0.0884\pm\sqrt{191.13101}}{-9.5428}=\frac{0.0884\pm 13.82501}{-9.5428}$

   The denominator is negative, so the plus branch produces the smaller root — a sign trap specific to downward parabolas.

4. 4

   Evaluate both branches.

   $x_1=\frac{0.0884+13.82501}{-9.5428}=-1.45800,\qquad x_2=\frac{0.0884-13.82501}{-9.5428}=1.43947$

5. 5

   Check with Vieta’s formulas. The sum of the roots must be $-\tfrac{b}{a}=-\tfrac{-0.0884}{-4.7714}=-0.018527$, and indeed $-1.45800+1.43947=-0.01853\;\checkmark$. The product must be $\tfrac{c}{a}=\tfrac{10.014}{-4.7714}=-2.09876$, and $(-1.45800)(1.43947)=-2.09875\;\checkmark$. Reporting the roots as $-1.4576$ and $1.4380$ would fail both checks in the fourth decimal.