(1/2)(2x+5)=7

Solution for (1/2)(2x+5)=7 equation:

(1/2)(2x+5)=7

We move all terms to the left:

(1/2)(2x+5)-(7)=0


Domain of the equation: 2)(2x+5)!=0

x∈R


We add all the numbers together, and all the variables

(+1/2)(2x+5)-7=0

We multiply parentheses ..

(+2x^2+1/2*5)-7=0

We multiply all the terms by the denominator


(+2x^2+1-7*2*5)=0

We get rid of parentheses

2x^2+1-7*2*5=0

We add all the numbers together, and all the variables

2x^2-69=0

a = 2; b = 0; c = -69;
Δ = b2-4ac
Δ = 02-4·2·(-69)
Δ = 552
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{552}=\sqrt{4*138}=\sqrt{4}*\sqrt{138}=2\sqrt{138}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{138}}{2*2}=\frac{0-2\sqrt{138}}{4}
=-\frac{2\sqrt{138}}{4}
=-\frac{\sqrt{138}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{138}}{2*2}=\frac{0+2\sqrt{138}}{4}
=\frac{2\sqrt{138}}{4}
=\frac{\sqrt{138}}{2}
$