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2x^2+48x+270=0
a = 2; b = 48; c = +270;
Δ = b2-4ac
Δ = 482-4·2·270
Δ = 144
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{144}=12$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(48)-12}{2*2}=\frac{-60}{4} =-15 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(48)+12}{2*2}=\frac{-36}{4} =-9 $
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