x = 15.3 or -0.3
Here, we want to get the value of x from the given equation
Let us bring like terms together, we have;

We now proceed to use the quadratic formula to solve this quadaratic equation. The formula is;
![x\text{ = }\frac{-b\text{ }\pm\text{ }\sqrt[]{b^2\text{ - 4ac}}}{2a}](https://img.qammunity.org/2023/formulas/mathematics/college/az0np1u8d52sprtutjbiz5w4rmpqgit2cr.png)
In the case of this question, a = 1 , b = -15 and c = -4
Substituting these values in the quadratic formula, we have;
![\begin{gathered} x\text{ = }\frac{-(-15)\text{ }\pm\text{ }\sqrt[]{(-15)^2\text{ - 4(1)(-4)}}}{2(1)} \\ \\ x\text{ = }\frac{15\text{ }\pm\text{ }\sqrt[]{225\text{ + 16}}}{2} \\ \\ x\text{ = }\frac{15\text{ }\pm\text{ }\sqrt[]{241}}{2} \\ \\ x\text{ = }\frac{15\text{ }\pm\text{ 15.52}}{2} \\ \\ x\text{ = }\frac{15\text{ + 15.52}}{2}\text{ or }(15-15.52)/(2) \\ \\ x\text{ = }15.26\text{ or -0.26} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/e26x1mazm4kzkiqzufk9fgabh5fyetrduc.png)
To the nearest tenth (one decimal place) x = 15.3 or -0.3